Showing that, if a pointwise-convergent sequence converges in norm in $\mathcal{C}^{2+\alpha}$, it does in $\mathcal{C}^{2+\beta}$ ($\beta < \alpha$)

Question

$ \newcommand{\CC}{\mathcal{C}} \newcommand{\a}{\alpha} \newcommand{\b}{\beta} \newcommand{\n}[1]{\left\| #1 \right\|} \newcommand{\nc}[2]{\left\| #1 \right\|_{\CC^{#2}(\oo)}} \newcommand{\seq}[1]{\left\{ #1_n \right\}_{n \in \mathbb{N}}} \newcommand{\o}{\Omega} \newcommand{\oo}{\overline{\Omega}} \newcommand{\ds}{\displaystyle} $Definitions: Some of the more useful and relevant definitions (e.g. $\alpha$-Hölder continuity, $\mathcal{C}^{2+\a}$, its norm, etc.) can be found in my answer here.

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Full Statement: Let $\o \subseteq \mathbb{R}^n$ (it may be bounded or unbounded, closed or open).

Consider a sequence $\seq u \subseteq \CC^{2+\a}(\oo)$ which is bounded above and pointwise convergent, i.e.

• $\exists M > 0$ where $\nc{u_n}{2+\a} < M$ (and this $M$ works for all $n$)
• $\ds \lim_{n \to \infty} u_n(x)$ exists, and we define $\ds u(x) := \lim_{n \to \infty} u_n(x)$

Let $0 < \beta < \alpha < 1$. We wish to show the following:

• $u \in \CC^{2+\beta}(\oo)$
• $\seq u$ converges in norm in $\CC^{2+\b}(\oo)$ specifically to $u$, i.e. $\nc{u_n-u}{2+\b} \xrightarrow{n\to\infty} 0$

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My Attempts and Understanding:

The issue of showing $u \in \CC^{2+\b}(\oo)$ has a bit of an issue for me. My thought process:

• Assume we can show that $u \in \CC^2(\oo)$. (This is another issue I have on recollection, showing it is $\CC^2$.)
• We can show that $u \in \CC^{2+\a}(\oo)$ then. This follows from continuity of norms and the boundedness of $\seq u$ fairly trivially.
• Since $u \in \CC^{2+\a}(\oo)$, we know it is twice differentiable and $\a$-Hölder continuous.
• We know that we may embed $\CC^\a(\oo)$ into $\CC^\b(\oo)$ for $0 < \b < \a < 1$.
• The result thus gives $u \in \CC^{2+\b}(\oo)$.

Now, as far showing the convergence in the norm of $\CC^{2+\b}$ (not $\CC^{2+\a}$), I'm completely lost. I've tried playing with more basic ideas (e.g. triangle inequality), but they're not tight enough to bound $\nc{u_n-u}{2+\b}$ by something that converges to $0$ (just by some constant). Playing around with the definitions of these (turning the norm into a sum of a bunch of suprema and derivatives) is very unenlightening as well.

So my questions:

• How might I prove $u \in \CC^2(\oo)$?
• How might I show the convergence in norm for $\CC^{2+\b}(\oo)$ (or, at least, are there any useful inequalities and properties you think I might find useful)?

Thanks for any help you might offer.

Answer 1

If $\Omega$ is bounded then the answer to your question is pretty much given by claim 2 here.

If $\Omega$ is unbounded then it isn't true. Take $u_n(x)=x^{\alpha+2}$, $x>0$. Then $u_n \in C^{2+\alpha}(\mathbb R_+)$ and $\| u_n\|_{C^{2+\alpha}(\mathbb R_+)}$ is constant but $u_n \not\in C^{2+\beta}(\mathbb R_+)$ for any $\beta \in (0,\alpha)$.