Prove that if $D^kf=0$ then $f$ is polynomial












0














Let $Esubseteqmathbb{R}^n$ an open connected set and let $f:Erightarrow mathbb{R}$ a $k+1$ times differentiable function such that $D^kf=0$. Show that $f$ is a polynomial in $x_1,...,x_n$ at a degree of at most $k$.





For any $piin{1,...,n}^n$ we're given that $frac{partial^{k+1}f}{partial x_{pi (1)}...partial x_{pi (k+1)}}=0$. Thus $frac{partial^{k}f}{partial x_{pi (1)}...partial x_{pi(k)}}=c_k$ and by induction $f=partial^0 f=x_{pi(1)}(x_{pi(2)}(...(x_{pi(k)}c_k+c_{k-1})...)+c_1)+c_0$ for some $c_iinmathbb{R}, forall 1leq i leq n$. I feel that this proof doesn't hold but I don't know where is the mistake and what is the right way. Thanks










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  • 3




    Taylor series might help?
    – copper.hat
    Nov 20 at 18:26










  • Your integration is false. I mean, for instance if $f(x,y)$ a function in two variables and $d^2 /dx dy f(x,y)=0$ then $d/dx f(x,y)= c(x)$ this is a function of $x$ and not just a constant. I think a proof by induction is more appropriate here.
    – Yanko
    Nov 20 at 18:28












  • You should not bring permutations into the game. The variable $k$ counting the number of differentiations is independent of the number $n$ of variables $x_i$.
    – Christian Blatter
    Nov 20 at 19:04










  • Such an $f$ is analytical and its series is finite...
    – Federico
    Nov 20 at 19:18










  • @Federico why $f$ is analytical or as you claim, entire?
    – Will M.
    Nov 20 at 19:19
















0














Let $Esubseteqmathbb{R}^n$ an open connected set and let $f:Erightarrow mathbb{R}$ a $k+1$ times differentiable function such that $D^kf=0$. Show that $f$ is a polynomial in $x_1,...,x_n$ at a degree of at most $k$.





For any $piin{1,...,n}^n$ we're given that $frac{partial^{k+1}f}{partial x_{pi (1)}...partial x_{pi (k+1)}}=0$. Thus $frac{partial^{k}f}{partial x_{pi (1)}...partial x_{pi(k)}}=c_k$ and by induction $f=partial^0 f=x_{pi(1)}(x_{pi(2)}(...(x_{pi(k)}c_k+c_{k-1})...)+c_1)+c_0$ for some $c_iinmathbb{R}, forall 1leq i leq n$. I feel that this proof doesn't hold but I don't know where is the mistake and what is the right way. Thanks










share|cite|improve this question




















  • 3




    Taylor series might help?
    – copper.hat
    Nov 20 at 18:26










  • Your integration is false. I mean, for instance if $f(x,y)$ a function in two variables and $d^2 /dx dy f(x,y)=0$ then $d/dx f(x,y)= c(x)$ this is a function of $x$ and not just a constant. I think a proof by induction is more appropriate here.
    – Yanko
    Nov 20 at 18:28












  • You should not bring permutations into the game. The variable $k$ counting the number of differentiations is independent of the number $n$ of variables $x_i$.
    – Christian Blatter
    Nov 20 at 19:04










  • Such an $f$ is analytical and its series is finite...
    – Federico
    Nov 20 at 19:18










  • @Federico why $f$ is analytical or as you claim, entire?
    – Will M.
    Nov 20 at 19:19














0












0








0







Let $Esubseteqmathbb{R}^n$ an open connected set and let $f:Erightarrow mathbb{R}$ a $k+1$ times differentiable function such that $D^kf=0$. Show that $f$ is a polynomial in $x_1,...,x_n$ at a degree of at most $k$.





For any $piin{1,...,n}^n$ we're given that $frac{partial^{k+1}f}{partial x_{pi (1)}...partial x_{pi (k+1)}}=0$. Thus $frac{partial^{k}f}{partial x_{pi (1)}...partial x_{pi(k)}}=c_k$ and by induction $f=partial^0 f=x_{pi(1)}(x_{pi(2)}(...(x_{pi(k)}c_k+c_{k-1})...)+c_1)+c_0$ for some $c_iinmathbb{R}, forall 1leq i leq n$. I feel that this proof doesn't hold but I don't know where is the mistake and what is the right way. Thanks










share|cite|improve this question















Let $Esubseteqmathbb{R}^n$ an open connected set and let $f:Erightarrow mathbb{R}$ a $k+1$ times differentiable function such that $D^kf=0$. Show that $f$ is a polynomial in $x_1,...,x_n$ at a degree of at most $k$.





For any $piin{1,...,n}^n$ we're given that $frac{partial^{k+1}f}{partial x_{pi (1)}...partial x_{pi (k+1)}}=0$. Thus $frac{partial^{k}f}{partial x_{pi (1)}...partial x_{pi(k)}}=c_k$ and by induction $f=partial^0 f=x_{pi(1)}(x_{pi(2)}(...(x_{pi(k)}c_k+c_{k-1})...)+c_1)+c_0$ for some $c_iinmathbb{R}, forall 1leq i leq n$. I feel that this proof doesn't hold but I don't know where is the mistake and what is the right way. Thanks







calculus derivatives partial-derivative






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edited Nov 20 at 18:26

























asked Nov 20 at 18:24









J. Doe

976




976








  • 3




    Taylor series might help?
    – copper.hat
    Nov 20 at 18:26










  • Your integration is false. I mean, for instance if $f(x,y)$ a function in two variables and $d^2 /dx dy f(x,y)=0$ then $d/dx f(x,y)= c(x)$ this is a function of $x$ and not just a constant. I think a proof by induction is more appropriate here.
    – Yanko
    Nov 20 at 18:28












  • You should not bring permutations into the game. The variable $k$ counting the number of differentiations is independent of the number $n$ of variables $x_i$.
    – Christian Blatter
    Nov 20 at 19:04










  • Such an $f$ is analytical and its series is finite...
    – Federico
    Nov 20 at 19:18










  • @Federico why $f$ is analytical or as you claim, entire?
    – Will M.
    Nov 20 at 19:19














  • 3




    Taylor series might help?
    – copper.hat
    Nov 20 at 18:26










  • Your integration is false. I mean, for instance if $f(x,y)$ a function in two variables and $d^2 /dx dy f(x,y)=0$ then $d/dx f(x,y)= c(x)$ this is a function of $x$ and not just a constant. I think a proof by induction is more appropriate here.
    – Yanko
    Nov 20 at 18:28












  • You should not bring permutations into the game. The variable $k$ counting the number of differentiations is independent of the number $n$ of variables $x_i$.
    – Christian Blatter
    Nov 20 at 19:04










  • Such an $f$ is analytical and its series is finite...
    – Federico
    Nov 20 at 19:18










  • @Federico why $f$ is analytical or as you claim, entire?
    – Will M.
    Nov 20 at 19:19








3




3




Taylor series might help?
– copper.hat
Nov 20 at 18:26




Taylor series might help?
– copper.hat
Nov 20 at 18:26












Your integration is false. I mean, for instance if $f(x,y)$ a function in two variables and $d^2 /dx dy f(x,y)=0$ then $d/dx f(x,y)= c(x)$ this is a function of $x$ and not just a constant. I think a proof by induction is more appropriate here.
– Yanko
Nov 20 at 18:28






Your integration is false. I mean, for instance if $f(x,y)$ a function in two variables and $d^2 /dx dy f(x,y)=0$ then $d/dx f(x,y)= c(x)$ this is a function of $x$ and not just a constant. I think a proof by induction is more appropriate here.
– Yanko
Nov 20 at 18:28














You should not bring permutations into the game. The variable $k$ counting the number of differentiations is independent of the number $n$ of variables $x_i$.
– Christian Blatter
Nov 20 at 19:04




You should not bring permutations into the game. The variable $k$ counting the number of differentiations is independent of the number $n$ of variables $x_i$.
– Christian Blatter
Nov 20 at 19:04












Such an $f$ is analytical and its series is finite...
– Federico
Nov 20 at 19:18




Such an $f$ is analytical and its series is finite...
– Federico
Nov 20 at 19:18












@Federico why $f$ is analytical or as you claim, entire?
– Will M.
Nov 20 at 19:19




@Federico why $f$ is analytical or as you claim, entire?
– Will M.
Nov 20 at 19:19










3 Answers
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Abridged solution. Proceed by induction on $k.$ The case $k = 0$ is very standard; consider a "base point" $a in E$ and the set $mathrm{X}$ of points in $x in mathrm{E}$ such that $f(x)=f(a).$ Then $mathrm{X}$ is closed in $mathrm{E}$ as $f$ is continuous. Now, $mathrm{X}$ is also open in $mathrm{E}$ for if $x in mathrm{X}$ then there is a small enough ball centred at $x$ contained in $mathrm{E}$ and for such ball the mean value theorem (applied to the line segment between $x$ and another point in the ball) gives that $f$ is constant in such ball, so such ball is subset of $mathrm{X}$ and then $mathrm{X}$ is open as claimed. Being $mathrm{E}$ connected, $mathrm{X} = mathrm{E}$ and the base case $k = 0$ is proved.



For the general case, assume $f$ satisfies $f^{(k+1)}=0$ on $mathrm{E}.$ The case $k = 0$ shows that $f^{(k)}$ is a constant $k$-linear function on $mathrm{E},$ call it $M.$ We show that $varphi:x mapsto f(x) - dfrac{1}{k!}M(x, ldots, x)$ satisfies $varphi^{(k)}=0$ on $mathrm{E}$ and hence the result. By standard results about derivatives of multilinear functions and the chain rule, we get at once that $varphi^{(k)}=f^{(k)}-dfrac{1}{k!} k!M = 0.$ Q.E.D.






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    0














    Note that $f$ is smooth since if $D^k f = 0$ then $D^l f = 0$ for $ l ge k$.



    Fix some $x_0 in E$, and suppose $C$ is an open convex set containing $x_0$ and contained in $E$. Then Taylor's theorem shows that if $y in C$ then we
    have $f(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$.



    Let $p(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$ be the polynomial, note that $p$ is defined everywhere and has maximum degree $k-1$. We wish to show that
    $f=p$ on $E$.



    As an aside, note that if $q$ is a polynomial and $q(x) = 0$ for $x$ in some open set then $q=0$.



    Pick some $x_nin E$ and note that since $E$ is open and connected there are $x_k in E$ such that the straight line curve $x_0 to x_1 to cdots to x_n$ is contained
    in $E$.



    Since the segment $[x_0,x_1] subset E$, there is some open convex set $C_1$ such that
    $[x_0,x_1] subset C_1 subset E$ and we have $f=p$ on $C_1$.



    Now let $p'$ be the Taylor expansion based at $x_1$. As above, there is some open
    convex set with $[x_1,x_2] subset C_2 subset E$ and we have $f=p'$ on $C_2$.



    Since $x_1 in C_1 cap C_2$, we see that $p=p'$ on an open set, and hence $p'=p$ everywhere.



    It is clear that we can repeat this process to see that $f(x_n) = p(x_n)$.



    Since $x_n in E$ was arbitrary, we see that $f=p$ on $E$.






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      -2














      Another alternative proof (which I guess is worng as well because I don't use the fact that $E$ is open and connected):



      From Taylor's formula we know that
      $$
      \ f(x_1,...,x_n)=sum_{i_1=1}^infty ...sum_{i_n=1}^infty frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}cdotfrac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)
      $$

      Let $0leq i_1,...,i_nleq k$. If $sum_{j=1}^n i_j>k$ then $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a)=0$. Otherwise, $frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}$ is polynom of a degree at most $k$, and $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)$ is a constant linear tranformation. Thus $f(x_1,...,x_n)$ is a polynom of degree at most $k$.






      share|cite|improve this answer





















      • Why $f$ is analytic?
        – Will M.
        Nov 20 at 19:18










      • I don't find the answre, why is it really? @WillM.
        – J. Doe
        Nov 20 at 19:32










      • I can only prove it is by well, showing it is a polynomial. This phenomenon of "a function is a polynomial provided one of its higher derivatives is zero identically" holds true in Banach spaces. As the proof I wrote shows.
        – Will M.
        Nov 20 at 19:56













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      3 Answers
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      Abridged solution. Proceed by induction on $k.$ The case $k = 0$ is very standard; consider a "base point" $a in E$ and the set $mathrm{X}$ of points in $x in mathrm{E}$ such that $f(x)=f(a).$ Then $mathrm{X}$ is closed in $mathrm{E}$ as $f$ is continuous. Now, $mathrm{X}$ is also open in $mathrm{E}$ for if $x in mathrm{X}$ then there is a small enough ball centred at $x$ contained in $mathrm{E}$ and for such ball the mean value theorem (applied to the line segment between $x$ and another point in the ball) gives that $f$ is constant in such ball, so such ball is subset of $mathrm{X}$ and then $mathrm{X}$ is open as claimed. Being $mathrm{E}$ connected, $mathrm{X} = mathrm{E}$ and the base case $k = 0$ is proved.



      For the general case, assume $f$ satisfies $f^{(k+1)}=0$ on $mathrm{E}.$ The case $k = 0$ shows that $f^{(k)}$ is a constant $k$-linear function on $mathrm{E},$ call it $M.$ We show that $varphi:x mapsto f(x) - dfrac{1}{k!}M(x, ldots, x)$ satisfies $varphi^{(k)}=0$ on $mathrm{E}$ and hence the result. By standard results about derivatives of multilinear functions and the chain rule, we get at once that $varphi^{(k)}=f^{(k)}-dfrac{1}{k!} k!M = 0.$ Q.E.D.






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        0














        Abridged solution. Proceed by induction on $k.$ The case $k = 0$ is very standard; consider a "base point" $a in E$ and the set $mathrm{X}$ of points in $x in mathrm{E}$ such that $f(x)=f(a).$ Then $mathrm{X}$ is closed in $mathrm{E}$ as $f$ is continuous. Now, $mathrm{X}$ is also open in $mathrm{E}$ for if $x in mathrm{X}$ then there is a small enough ball centred at $x$ contained in $mathrm{E}$ and for such ball the mean value theorem (applied to the line segment between $x$ and another point in the ball) gives that $f$ is constant in such ball, so such ball is subset of $mathrm{X}$ and then $mathrm{X}$ is open as claimed. Being $mathrm{E}$ connected, $mathrm{X} = mathrm{E}$ and the base case $k = 0$ is proved.



        For the general case, assume $f$ satisfies $f^{(k+1)}=0$ on $mathrm{E}.$ The case $k = 0$ shows that $f^{(k)}$ is a constant $k$-linear function on $mathrm{E},$ call it $M.$ We show that $varphi:x mapsto f(x) - dfrac{1}{k!}M(x, ldots, x)$ satisfies $varphi^{(k)}=0$ on $mathrm{E}$ and hence the result. By standard results about derivatives of multilinear functions and the chain rule, we get at once that $varphi^{(k)}=f^{(k)}-dfrac{1}{k!} k!M = 0.$ Q.E.D.






        share|cite|improve this answer
























          0












          0








          0






          Abridged solution. Proceed by induction on $k.$ The case $k = 0$ is very standard; consider a "base point" $a in E$ and the set $mathrm{X}$ of points in $x in mathrm{E}$ such that $f(x)=f(a).$ Then $mathrm{X}$ is closed in $mathrm{E}$ as $f$ is continuous. Now, $mathrm{X}$ is also open in $mathrm{E}$ for if $x in mathrm{X}$ then there is a small enough ball centred at $x$ contained in $mathrm{E}$ and for such ball the mean value theorem (applied to the line segment between $x$ and another point in the ball) gives that $f$ is constant in such ball, so such ball is subset of $mathrm{X}$ and then $mathrm{X}$ is open as claimed. Being $mathrm{E}$ connected, $mathrm{X} = mathrm{E}$ and the base case $k = 0$ is proved.



          For the general case, assume $f$ satisfies $f^{(k+1)}=0$ on $mathrm{E}.$ The case $k = 0$ shows that $f^{(k)}$ is a constant $k$-linear function on $mathrm{E},$ call it $M.$ We show that $varphi:x mapsto f(x) - dfrac{1}{k!}M(x, ldots, x)$ satisfies $varphi^{(k)}=0$ on $mathrm{E}$ and hence the result. By standard results about derivatives of multilinear functions and the chain rule, we get at once that $varphi^{(k)}=f^{(k)}-dfrac{1}{k!} k!M = 0.$ Q.E.D.






          share|cite|improve this answer












          Abridged solution. Proceed by induction on $k.$ The case $k = 0$ is very standard; consider a "base point" $a in E$ and the set $mathrm{X}$ of points in $x in mathrm{E}$ such that $f(x)=f(a).$ Then $mathrm{X}$ is closed in $mathrm{E}$ as $f$ is continuous. Now, $mathrm{X}$ is also open in $mathrm{E}$ for if $x in mathrm{X}$ then there is a small enough ball centred at $x$ contained in $mathrm{E}$ and for such ball the mean value theorem (applied to the line segment between $x$ and another point in the ball) gives that $f$ is constant in such ball, so such ball is subset of $mathrm{X}$ and then $mathrm{X}$ is open as claimed. Being $mathrm{E}$ connected, $mathrm{X} = mathrm{E}$ and the base case $k = 0$ is proved.



          For the general case, assume $f$ satisfies $f^{(k+1)}=0$ on $mathrm{E}.$ The case $k = 0$ shows that $f^{(k)}$ is a constant $k$-linear function on $mathrm{E},$ call it $M.$ We show that $varphi:x mapsto f(x) - dfrac{1}{k!}M(x, ldots, x)$ satisfies $varphi^{(k)}=0$ on $mathrm{E}$ and hence the result. By standard results about derivatives of multilinear functions and the chain rule, we get at once that $varphi^{(k)}=f^{(k)}-dfrac{1}{k!} k!M = 0.$ Q.E.D.







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          share|cite|improve this answer










          answered Nov 20 at 19:33









          Will M.

          2,377314




          2,377314























              0














              Note that $f$ is smooth since if $D^k f = 0$ then $D^l f = 0$ for $ l ge k$.



              Fix some $x_0 in E$, and suppose $C$ is an open convex set containing $x_0$ and contained in $E$. Then Taylor's theorem shows that if $y in C$ then we
              have $f(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$.



              Let $p(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$ be the polynomial, note that $p$ is defined everywhere and has maximum degree $k-1$. We wish to show that
              $f=p$ on $E$.



              As an aside, note that if $q$ is a polynomial and $q(x) = 0$ for $x$ in some open set then $q=0$.



              Pick some $x_nin E$ and note that since $E$ is open and connected there are $x_k in E$ such that the straight line curve $x_0 to x_1 to cdots to x_n$ is contained
              in $E$.



              Since the segment $[x_0,x_1] subset E$, there is some open convex set $C_1$ such that
              $[x_0,x_1] subset C_1 subset E$ and we have $f=p$ on $C_1$.



              Now let $p'$ be the Taylor expansion based at $x_1$. As above, there is some open
              convex set with $[x_1,x_2] subset C_2 subset E$ and we have $f=p'$ on $C_2$.



              Since $x_1 in C_1 cap C_2$, we see that $p=p'$ on an open set, and hence $p'=p$ everywhere.



              It is clear that we can repeat this process to see that $f(x_n) = p(x_n)$.



              Since $x_n in E$ was arbitrary, we see that $f=p$ on $E$.






              share|cite|improve this answer


























                0














                Note that $f$ is smooth since if $D^k f = 0$ then $D^l f = 0$ for $ l ge k$.



                Fix some $x_0 in E$, and suppose $C$ is an open convex set containing $x_0$ and contained in $E$. Then Taylor's theorem shows that if $y in C$ then we
                have $f(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$.



                Let $p(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$ be the polynomial, note that $p$ is defined everywhere and has maximum degree $k-1$. We wish to show that
                $f=p$ on $E$.



                As an aside, note that if $q$ is a polynomial and $q(x) = 0$ for $x$ in some open set then $q=0$.



                Pick some $x_nin E$ and note that since $E$ is open and connected there are $x_k in E$ such that the straight line curve $x_0 to x_1 to cdots to x_n$ is contained
                in $E$.



                Since the segment $[x_0,x_1] subset E$, there is some open convex set $C_1$ such that
                $[x_0,x_1] subset C_1 subset E$ and we have $f=p$ on $C_1$.



                Now let $p'$ be the Taylor expansion based at $x_1$. As above, there is some open
                convex set with $[x_1,x_2] subset C_2 subset E$ and we have $f=p'$ on $C_2$.



                Since $x_1 in C_1 cap C_2$, we see that $p=p'$ on an open set, and hence $p'=p$ everywhere.



                It is clear that we can repeat this process to see that $f(x_n) = p(x_n)$.



                Since $x_n in E$ was arbitrary, we see that $f=p$ on $E$.






                share|cite|improve this answer
























                  0












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                  0






                  Note that $f$ is smooth since if $D^k f = 0$ then $D^l f = 0$ for $ l ge k$.



                  Fix some $x_0 in E$, and suppose $C$ is an open convex set containing $x_0$ and contained in $E$. Then Taylor's theorem shows that if $y in C$ then we
                  have $f(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$.



                  Let $p(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$ be the polynomial, note that $p$ is defined everywhere and has maximum degree $k-1$. We wish to show that
                  $f=p$ on $E$.



                  As an aside, note that if $q$ is a polynomial and $q(x) = 0$ for $x$ in some open set then $q=0$.



                  Pick some $x_nin E$ and note that since $E$ is open and connected there are $x_k in E$ such that the straight line curve $x_0 to x_1 to cdots to x_n$ is contained
                  in $E$.



                  Since the segment $[x_0,x_1] subset E$, there is some open convex set $C_1$ such that
                  $[x_0,x_1] subset C_1 subset E$ and we have $f=p$ on $C_1$.



                  Now let $p'$ be the Taylor expansion based at $x_1$. As above, there is some open
                  convex set with $[x_1,x_2] subset C_2 subset E$ and we have $f=p'$ on $C_2$.



                  Since $x_1 in C_1 cap C_2$, we see that $p=p'$ on an open set, and hence $p'=p$ everywhere.



                  It is clear that we can repeat this process to see that $f(x_n) = p(x_n)$.



                  Since $x_n in E$ was arbitrary, we see that $f=p$ on $E$.






                  share|cite|improve this answer












                  Note that $f$ is smooth since if $D^k f = 0$ then $D^l f = 0$ for $ l ge k$.



                  Fix some $x_0 in E$, and suppose $C$ is an open convex set containing $x_0$ and contained in $E$. Then Taylor's theorem shows that if $y in C$ then we
                  have $f(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$.



                  Let $p(y) = sum_{|alpha| le k-1} {1 over alpha !} D^alpha f(x_0) (y-x_0)^alpha$ be the polynomial, note that $p$ is defined everywhere and has maximum degree $k-1$. We wish to show that
                  $f=p$ on $E$.



                  As an aside, note that if $q$ is a polynomial and $q(x) = 0$ for $x$ in some open set then $q=0$.



                  Pick some $x_nin E$ and note that since $E$ is open and connected there are $x_k in E$ such that the straight line curve $x_0 to x_1 to cdots to x_n$ is contained
                  in $E$.



                  Since the segment $[x_0,x_1] subset E$, there is some open convex set $C_1$ such that
                  $[x_0,x_1] subset C_1 subset E$ and we have $f=p$ on $C_1$.



                  Now let $p'$ be the Taylor expansion based at $x_1$. As above, there is some open
                  convex set with $[x_1,x_2] subset C_2 subset E$ and we have $f=p'$ on $C_2$.



                  Since $x_1 in C_1 cap C_2$, we see that $p=p'$ on an open set, and hence $p'=p$ everywhere.



                  It is clear that we can repeat this process to see that $f(x_n) = p(x_n)$.



                  Since $x_n in E$ was arbitrary, we see that $f=p$ on $E$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 4:21









                  copper.hat

                  126k559159




                  126k559159























                      -2














                      Another alternative proof (which I guess is worng as well because I don't use the fact that $E$ is open and connected):



                      From Taylor's formula we know that
                      $$
                      \ f(x_1,...,x_n)=sum_{i_1=1}^infty ...sum_{i_n=1}^infty frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}cdotfrac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)
                      $$

                      Let $0leq i_1,...,i_nleq k$. If $sum_{j=1}^n i_j>k$ then $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a)=0$. Otherwise, $frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}$ is polynom of a degree at most $k$, and $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)$ is a constant linear tranformation. Thus $f(x_1,...,x_n)$ is a polynom of degree at most $k$.






                      share|cite|improve this answer





















                      • Why $f$ is analytic?
                        – Will M.
                        Nov 20 at 19:18










                      • I don't find the answre, why is it really? @WillM.
                        – J. Doe
                        Nov 20 at 19:32










                      • I can only prove it is by well, showing it is a polynomial. This phenomenon of "a function is a polynomial provided one of its higher derivatives is zero identically" holds true in Banach spaces. As the proof I wrote shows.
                        – Will M.
                        Nov 20 at 19:56


















                      -2














                      Another alternative proof (which I guess is worng as well because I don't use the fact that $E$ is open and connected):



                      From Taylor's formula we know that
                      $$
                      \ f(x_1,...,x_n)=sum_{i_1=1}^infty ...sum_{i_n=1}^infty frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}cdotfrac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)
                      $$

                      Let $0leq i_1,...,i_nleq k$. If $sum_{j=1}^n i_j>k$ then $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a)=0$. Otherwise, $frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}$ is polynom of a degree at most $k$, and $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)$ is a constant linear tranformation. Thus $f(x_1,...,x_n)$ is a polynom of degree at most $k$.






                      share|cite|improve this answer





















                      • Why $f$ is analytic?
                        – Will M.
                        Nov 20 at 19:18










                      • I don't find the answre, why is it really? @WillM.
                        – J. Doe
                        Nov 20 at 19:32










                      • I can only prove it is by well, showing it is a polynomial. This phenomenon of "a function is a polynomial provided one of its higher derivatives is zero identically" holds true in Banach spaces. As the proof I wrote shows.
                        – Will M.
                        Nov 20 at 19:56
















                      -2












                      -2








                      -2






                      Another alternative proof (which I guess is worng as well because I don't use the fact that $E$ is open and connected):



                      From Taylor's formula we know that
                      $$
                      \ f(x_1,...,x_n)=sum_{i_1=1}^infty ...sum_{i_n=1}^infty frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}cdotfrac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)
                      $$

                      Let $0leq i_1,...,i_nleq k$. If $sum_{j=1}^n i_j>k$ then $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a)=0$. Otherwise, $frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}$ is polynom of a degree at most $k$, and $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)$ is a constant linear tranformation. Thus $f(x_1,...,x_n)$ is a polynom of degree at most $k$.






                      share|cite|improve this answer












                      Another alternative proof (which I guess is worng as well because I don't use the fact that $E$ is open and connected):



                      From Taylor's formula we know that
                      $$
                      \ f(x_1,...,x_n)=sum_{i_1=1}^infty ...sum_{i_n=1}^infty frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}cdotfrac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)
                      $$

                      Let $0leq i_1,...,i_nleq k$. If $sum_{j=1}^n i_j>k$ then $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a)=0$. Otherwise, $frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}$ is polynom of a degree at most $k$, and $frac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)$ is a constant linear tranformation. Thus $f(x_1,...,x_n)$ is a polynom of degree at most $k$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 20 at 19:11









                      J. Doe

                      976




                      976












                      • Why $f$ is analytic?
                        – Will M.
                        Nov 20 at 19:18










                      • I don't find the answre, why is it really? @WillM.
                        – J. Doe
                        Nov 20 at 19:32










                      • I can only prove it is by well, showing it is a polynomial. This phenomenon of "a function is a polynomial provided one of its higher derivatives is zero identically" holds true in Banach spaces. As the proof I wrote shows.
                        – Will M.
                        Nov 20 at 19:56




















                      • Why $f$ is analytic?
                        – Will M.
                        Nov 20 at 19:18










                      • I don't find the answre, why is it really? @WillM.
                        – J. Doe
                        Nov 20 at 19:32










                      • I can only prove it is by well, showing it is a polynomial. This phenomenon of "a function is a polynomial provided one of its higher derivatives is zero identically" holds true in Banach spaces. As the proof I wrote shows.
                        – Will M.
                        Nov 20 at 19:56


















                      Why $f$ is analytic?
                      – Will M.
                      Nov 20 at 19:18




                      Why $f$ is analytic?
                      – Will M.
                      Nov 20 at 19:18












                      I don't find the answre, why is it really? @WillM.
                      – J. Doe
                      Nov 20 at 19:32




                      I don't find the answre, why is it really? @WillM.
                      – J. Doe
                      Nov 20 at 19:32












                      I can only prove it is by well, showing it is a polynomial. This phenomenon of "a function is a polynomial provided one of its higher derivatives is zero identically" holds true in Banach spaces. As the proof I wrote shows.
                      – Will M.
                      Nov 20 at 19:56






                      I can only prove it is by well, showing it is a polynomial. This phenomenon of "a function is a polynomial provided one of its higher derivatives is zero identically" holds true in Banach spaces. As the proof I wrote shows.
                      – Will M.
                      Nov 20 at 19:56




















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