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Let $E_i$ be a normed $\mathbb R$-vector space and $\Omega\subseteq E_1$ be open. We can easily show that if $f:\Omega\to E_2$ be Fréchet differentiable

  1. $\Omega$ is convex and ${\rm D}f$ is bounded$^1$; or
  2. $\Omega$ is compact$^2$,

then $f\in C^{0,\:1}(\Omega,E_2)$.$^3$

Now I've read that if $E_1=\mathbb R^d$, for some $d\in\mathbb N$, $E_2=\mathbb R$ and $f\in C_c^\infty(\Omega)$, then $f\in C^{0,\:\alpha}(\overline\Omega)$ for all $\alpha\in(0,1]$.

How can we prove this? And does this conclusion hold in more general situations as well?

$^1$ In that case, the Hahn-Banach theorem implies $$\forall x,y\in\Omega:\left\|f(x)-f(y)\right\|_{E_2}\le\sup_{z\in\Omega}\left\|{\rm D}f(z)\right\|_{\mathfrak L(E_1,\:E_2)}\left\|x-y\right\|_{E_1}.\tag2$$

$^2$ In that case, the claim can be shown as described in this answer.

$^3$ Remember that if $(M_i,d_i)$ is a metric space and $\alpha\in(0,1]$, then $$\left\|g\right\|_{C^{0,\:\alpha}(M_1,\:M_2)}:=\sup_{\substack{x,\:y\:\in\:M_1\\x\:\ne\:y}}\frac{d_2(g(x),g(y))}{d_1(x,y)^\alpha}\;\;\;\text{for }g:M_1\to M_2$$ and $$C^{0,\:\alpha}(M_1,M_2):=\left\{g:M_1\to M_2\mid\left\|g\right\|_{C^{0,\:\alpha}(M_1,\:M_2)}<\infty\right\}.$$

0xbadf00d
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  • your norm on $C^{0,\alpha}$ is written wrongly, with both $f$ and $g$ used, and without any reference to $\alpha$. The shortest way to go is probably to show $C^{0,1}(\Omega)\subset C^{0,\alpha}(\Omega)$ – Calvin Khor Jun 16 '21 at 08:48
  • @CalvinKhor Thank you for your comment. I've fixed the typos. I know that $C^{0,:1}(M_1,M_2)\subseteq C^{0,:\alpha}(\Omega)$, but how can we show that $f\in C^{0,:\alpha}(\overline\Omega)$ (i.e. that there is an Hölder continuous extension to the closure)? – 0xbadf00d Jun 17 '21 at 05:00
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    The same proof for uniform continuity. Let $a$ be a boundary point and $a_n\to a$ from the interior. Then $f(a_n)$ is Cauchy so we can define a candidate function on the closure. Then $|f(a_n) - f(b) |\lesssim |a_n - b|^\alpha $ and taking limits gives the required Holder estimate at $a$ for $b$ not in the boundary, and a similar calc works if you are checking the semi norm along the boundary. It still works if you change to a different modulus of continuity and use metrics instead of norms – Calvin Khor Jun 17 '21 at 05:23
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    @CalvinKhor Thank you for your comment. For completeness, I've provided an answer. Do you agree? – 0xbadf00d Jun 17 '21 at 19:23
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    Yes, except for a small typo after the limit you still have $x_n ,y_n$ and alpha disappeared – Calvin Khor Jun 17 '21 at 22:58
  • @CalvinKhor Thank your for the confirmation. I've got a related question: https://math.stackexchange.com/q/4176186/47771. Would be great if you could take a look at it. – 0xbadf00d Jun 18 '21 at 05:00

1 Answers1

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Let $(M_1,d_1)$ be a metric space, $\Omega\subseteq M_1$, $(M_2,d_2)$ be a complete metric space, $\alpha\in(0,1]$ and $f\in C^{0,\:\alpha}(\Omega,M_2)$.

By the defining property of Hölder continuity, it is easy to see that if $(x_n)_{n\in\mathbb N}\subseteq\Omega$ is Cauchy, then $(f(x_n))_{n\in\mathbb N}$ is Cauchy as well. Since $(M_2,d_2)$ is complete, this implies that there is an unique $\overline f\in C(\overline\Omega,E_2)$ with $\left.\overline f\right|_\Omega=f$.

Let $x,y\in\overline\Omega$. Then there are $(x_n)_{n\in\mathbb N},(y_n)_{n\in\mathbb N}\subseteq\Omega$ with $$d_1(x_n,x)+d_1(y_n,y)\xrightarrow{n\to\infty}0\tag3$$ and hence (by continuity of $d_1$ and $d_2$) \begin{equation}\begin{split}d_2\left(\overline f(x),\overline f(y)\right)&=\lim_{n\to\infty}d_2(f(x_n),f(y_n))\\&\le\left\|f\right\|_{C^{0,\:\alpha}(\Omega,\:M_2)}\lim_{n\to\infty}d_1(x_n,y_n)^\alpha=\left\|f\right\|_{C^{0,\:\alpha}(\Omega,\:M_2)}d_1(x,y).\end{split}\tag4\end{equation}

We have shown that each function in $C^{0,\:\alpha}(\Omega,M_2)$ has a unique extension to a function in $C^{0,\:\alpha}\left(\overline\Omega,M_2\right)$ and that the induced embedding of $C^{0,\:\alpha}(\Omega,M_2)$ into $C^{0,\:\alpha}\left(\overline\Omega,M_2\right)$ is continuous.

0xbadf00d
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