Holder continuity of the derivate of $|x|^alpha$ for $alpha>1$











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Suppose $U=B(0,1)$ is an open ball in $mathbb R^m$ and $alpha>1$. My question is if $|x|^alphain C^{1,gamma}(U)$ for some $gammain (0,1]$. I know that $|x|^alphain C^1(U)$ and
$$
f_i(x):=frac{partial}{partial x_i}|x|^alpha=alpha|x|^{alpha-2}x_i, mbox{ if }xneq 0.
$$
If $alphageq 2$, it is easy because we obtain $|x|^alphain C^2(U)subset C^{1,1}(U)$, with $gamma=1$ in this case.



We can suppose $1<alpha<2$. If $|x|^alphain C^{1,gamma}(U)$ then using $x_n=frac{e_i}n$ and $y_n=-frac{e_i}n$ we obtain
$$
frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^gamma}=frac{alpha}{2^{gamma-1}}n^{gamma-(alpha-1)}overset{nrightarrow+infty}longrightarrow+infty,
$$
if $gamma>alpha-1$. If this $gamma$ exist we must have $gammaleqalpha-1$. I guess that $|x|^alphain C^{1,alpha-1}(U)$.



I try proof that exist $C>0$ such that
$$
|f_i(x)-f_i(y)|leq C|x-y|^{alpha-1},
$$
but I only succeeded in the case $m=1$.






functional-analysis analysis inequality pde







share|cite|improve this question


























    up vote
    1
    down vote

    favorite












    Suppose $U=B(0,1)$ is an open ball in $mathbb R^m$ and $alpha>1$. My question is if $|x|^alphain C^{1,gamma}(U)$ for some $gammain (0,1]$. I know that $|x|^alphain C^1(U)$ and
    $$
    f_i(x):=frac{partial}{partial x_i}|x|^alpha=alpha|x|^{alpha-2}x_i, mbox{ if }xneq 0.
    $$
    If $alphageq 2$, it is easy because we obtain $|x|^alphain C^2(U)subset C^{1,1}(U)$, with $gamma=1$ in this case.



    We can suppose $1<alpha<2$. If $|x|^alphain C^{1,gamma}(U)$ then using $x_n=frac{e_i}n$ and $y_n=-frac{e_i}n$ we obtain
    $$
    frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^gamma}=frac{alpha}{2^{gamma-1}}n^{gamma-(alpha-1)}overset{nrightarrow+infty}longrightarrow+infty,
    $$
    if $gamma>alpha-1$. If this $gamma$ exist we must have $gammaleqalpha-1$. I guess that $|x|^alphain C^{1,alpha-1}(U)$.



    I try proof that exist $C>0$ such that
    $$
    |f_i(x)-f_i(y)|leq C|x-y|^{alpha-1},
    $$
    but I only succeeded in the case $m=1$.






    functional-analysis analysis inequality pde







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Suppose $U=B(0,1)$ is an open ball in $mathbb R^m$ and $alpha>1$. My question is if $|x|^alphain C^{1,gamma}(U)$ for some $gammain (0,1]$. I know that $|x|^alphain C^1(U)$ and
      $$
      f_i(x):=frac{partial}{partial x_i}|x|^alpha=alpha|x|^{alpha-2}x_i, mbox{ if }xneq 0.
      $$
      If $alphageq 2$, it is easy because we obtain $|x|^alphain C^2(U)subset C^{1,1}(U)$, with $gamma=1$ in this case.



      We can suppose $1<alpha<2$. If $|x|^alphain C^{1,gamma}(U)$ then using $x_n=frac{e_i}n$ and $y_n=-frac{e_i}n$ we obtain
      $$
      frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^gamma}=frac{alpha}{2^{gamma-1}}n^{gamma-(alpha-1)}overset{nrightarrow+infty}longrightarrow+infty,
      $$
      if $gamma>alpha-1$. If this $gamma$ exist we must have $gammaleqalpha-1$. I guess that $|x|^alphain C^{1,alpha-1}(U)$.



      I try proof that exist $C>0$ such that
      $$
      |f_i(x)-f_i(y)|leq C|x-y|^{alpha-1},
      $$
      but I only succeeded in the case $m=1$.






      functional-analysis analysis inequality pde







      share|cite|improve this question













      Suppose $U=B(0,1)$ is an open ball in $mathbb R^m$ and $alpha>1$. My question is if $|x|^alphain C^{1,gamma}(U)$ for some $gammain (0,1]$. I know that $|x|^alphain C^1(U)$ and
      $$
      f_i(x):=frac{partial}{partial x_i}|x|^alpha=alpha|x|^{alpha-2}x_i, mbox{ if }xneq 0.
      $$
      If $alphageq 2$, it is easy because we obtain $|x|^alphain C^2(U)subset C^{1,1}(U)$, with $gamma=1$ in this case.



      We can suppose $1<alpha<2$. If $|x|^alphain C^{1,gamma}(U)$ then using $x_n=frac{e_i}n$ and $y_n=-frac{e_i}n$ we obtain
      $$
      frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^gamma}=frac{alpha}{2^{gamma-1}}n^{gamma-(alpha-1)}overset{nrightarrow+infty}longrightarrow+infty,
      $$
      if $gamma>alpha-1$. If this $gamma$ exist we must have $gammaleqalpha-1$. I guess that $|x|^alphain C^{1,alpha-1}(U)$.



      I try proof that exist $C>0$ such that
      $$
      |f_i(x)-f_i(y)|leq C|x-y|^{alpha-1},
      $$
      but I only succeeded in the case $m=1$.






      functional-analysis analysis inequality pde




      functional-analysis analysis inequality pde






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 19 at 21:12









      Raoní Cabral Ponciano

      759




      759






















          1 Answer
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          First, let me prove this for $m=1$. The question is, whether the function
          $$
          g(x):= sign(x)|x|^gamma
          $$
          is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
          $$
          g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
          $$
          and by Hoelder's inequality
          $$
          |g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
          = gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
          le gamma (x-y)^gamma .
          $$
          In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
          $$
          frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
          $$
          This shows $gin C^gamma([-1,1])$.



          Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
          $$
          f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
          $$
          The second term can be reduced to the case $m=1$. The first term is unproblematic
          if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
          Then
          $$
          left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
          le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
          $$
          This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .






          share|cite|improve this answer





















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            up vote
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            down vote



            accepted










            First, let me prove this for $m=1$. The question is, whether the function
            $$
            g(x):= sign(x)|x|^gamma
            $$
            is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
            $$
            g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
            $$
            and by Hoelder's inequality
            $$
            |g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
            = gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
            le gamma (x-y)^gamma .
            $$
            In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
            $$
            frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
            $$
            This shows $gin C^gamma([-1,1])$.



            Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
            $$
            f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
            $$
            The second term can be reduced to the case $m=1$. The first term is unproblematic
            if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
            Then
            $$
            left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
            le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
            $$
            This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              First, let me prove this for $m=1$. The question is, whether the function
              $$
              g(x):= sign(x)|x|^gamma
              $$
              is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
              $$
              g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
              $$
              and by Hoelder's inequality
              $$
              |g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
              = gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
              le gamma (x-y)^gamma .
              $$
              In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
              $$
              frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
              $$
              This shows $gin C^gamma([-1,1])$.



              Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
              $$
              f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
              $$
              The second term can be reduced to the case $m=1$. The first term is unproblematic
              if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
              Then
              $$
              left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
              le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
              $$
              This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                First, let me prove this for $m=1$. The question is, whether the function
                $$
                g(x):= sign(x)|x|^gamma
                $$
                is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
                $$
                g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
                $$
                and by Hoelder's inequality
                $$
                |g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
                = gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
                le gamma (x-y)^gamma .
                $$
                In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
                $$
                frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
                $$
                This shows $gin C^gamma([-1,1])$.



                Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
                $$
                f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
                $$
                The second term can be reduced to the case $m=1$. The first term is unproblematic
                if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
                Then
                $$
                left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
                le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
                $$
                This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .






                share|cite|improve this answer












                First, let me prove this for $m=1$. The question is, whether the function
                $$
                g(x):= sign(x)|x|^gamma
                $$
                is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
                $$
                g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
                $$
                and by Hoelder's inequality
                $$
                |g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
                = gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
                le gamma (x-y)^gamma .
                $$
                In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
                $$
                frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
                $$
                This shows $gin C^gamma([-1,1])$.



                Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
                $$
                f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
                $$
                The second term can be reduced to the case $m=1$. The first term is unproblematic
                if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
                Then
                $$
                left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
                le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
                $$
                This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .







                share|cite|improve this answer












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                answered Nov 20 at 8:03









                daw

                24k1544




                24k1544






























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