How can a discontinuous function belong to $C_B^1(Omega)$, the space of continuous functions $u$ with bounded...












0














Let $Omega = {(x,y) in mathbb{R}^2 : 0 < |x| < 1, 0 < y < 1}$ and consider the function $u$ defined on $Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10)
$$
u(x,y) =
begin{cases}
1, quad x > 0, \
0, quad x < 0.
end{cases}
$$



On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(Omega)$ which consists of function in $C^1(Omega)$ such that $D^alpha u$ is bounded for $0 le alpha le 1$.



But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?






pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions







share|cite|improve this question



























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    Let $Omega = {(x,y) in mathbb{R}^2 : 0 < |x| < 1, 0 < y < 1}$ and consider the function $u$ defined on $Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10)
    $$
    u(x,y) =
    begin{cases}
    1, quad x > 0, \
    0, quad x < 0.
    end{cases}
    $$



    On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(Omega)$ which consists of function in $C^1(Omega)$ such that $D^alpha u$ is bounded for $0 le alpha le 1$.



    But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?






    pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions







    share|cite|improve this question

























      0












      0








      0


      1





      Let $Omega = {(x,y) in mathbb{R}^2 : 0 < |x| < 1, 0 < y < 1}$ and consider the function $u$ defined on $Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10)
      $$
      u(x,y) =
      begin{cases}
      1, quad x > 0, \
      0, quad x < 0.
      end{cases}
      $$



      On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(Omega)$ which consists of function in $C^1(Omega)$ such that $D^alpha u$ is bounded for $0 le alpha le 1$.



      But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?






      pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions







      share|cite|improve this question













      Let $Omega = {(x,y) in mathbb{R}^2 : 0 < |x| < 1, 0 < y < 1}$ and consider the function $u$ defined on $Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10)
      $$
      u(x,y) =
      begin{cases}
      1, quad x > 0, \
      0, quad x < 0.
      end{cases}
      $$



      On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(Omega)$ which consists of function in $C^1(Omega)$ such that $D^alpha u$ is bounded for $0 le alpha le 1$.



      But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?






      pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions




      pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 21 '18 at 10:58









      eurocoder

      1,095315




      1,095315






















          1 Answer
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          Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
          $$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
          $f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).






          share|cite|improve this answer





















          • Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
            – eurocoder
            Nov 21 '18 at 11:23












          • Nevermind, he says on page that 10 that this is an abuse of notation.
            – eurocoder
            Nov 21 '18 at 11:26










          • If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
            – Calvin Khor
            Nov 21 '18 at 11:26













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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
          $$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
          $f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).






          share|cite|improve this answer





















          • Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
            – eurocoder
            Nov 21 '18 at 11:23












          • Nevermind, he says on page that 10 that this is an abuse of notation.
            – eurocoder
            Nov 21 '18 at 11:26










          • If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
            – Calvin Khor
            Nov 21 '18 at 11:26


















          3














          Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
          $$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
          $f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).






          share|cite|improve this answer





















          • Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
            – eurocoder
            Nov 21 '18 at 11:23












          • Nevermind, he says on page that 10 that this is an abuse of notation.
            – eurocoder
            Nov 21 '18 at 11:26










          • If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
            – Calvin Khor
            Nov 21 '18 at 11:26
















          3












          3








          3






          Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
          $$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
          $f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).






          share|cite|improve this answer












          Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
          $$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
          $f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 '18 at 11:18









          Calvin Khor

          11.2k21438




          11.2k21438












          • Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
            – eurocoder
            Nov 21 '18 at 11:23












          • Nevermind, he says on page that 10 that this is an abuse of notation.
            – eurocoder
            Nov 21 '18 at 11:26










          • If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
            – Calvin Khor
            Nov 21 '18 at 11:26




















          • Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
            – eurocoder
            Nov 21 '18 at 11:23












          • Nevermind, he says on page that 10 that this is an abuse of notation.
            – eurocoder
            Nov 21 '18 at 11:26










          • If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
            – Calvin Khor
            Nov 21 '18 at 11:26


















          Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
          – eurocoder
          Nov 21 '18 at 11:23






          Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
          – eurocoder
          Nov 21 '18 at 11:23














          Nevermind, he says on page that 10 that this is an abuse of notation.
          – eurocoder
          Nov 21 '18 at 11:26




          Nevermind, he says on page that 10 that this is an abuse of notation.
          – eurocoder
          Nov 21 '18 at 11:26












          If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
          – Calvin Khor
          Nov 21 '18 at 11:26






          If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
          – Calvin Khor
          Nov 21 '18 at 11:26




















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