Tagged Questions

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Decay of Fourier Coefficients implies Holder Continuity?

This is an exercise problem. I got stuck here and would like to get a hint. The problem is Suppose $f$ is continuous and $2\pi$-periodic, and $|\hat{f}(n)|\leq |n|^{-3/2}$ for all non-zero ...
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Showing a set is nowhere dense in $C([0,1])$

Let $E_n$ be the set of all $f \in C\big([0,1]\big)$ for which there exists $x_0 \in [0,1]$ (depending on $f$) such that \begin{align*} \lvert\, f(x)-f(x_0)\rvert \leq n\lvert x-x_0\rvert, ...
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Is $||u||_{C^\alpha} \leq ||u||_{C^1}$ for all $u$?

We have $||u||_{C^\alpha,\Omega} = \text{sup}_\Omega |u(x)|+ \text{sup}_\Omega \frac{|u(x)-u(y)|}{|x-y|^\alpha}$ and $||u||_{C^1} =\text{sup}_\Omega |u(x)| + \text{sup}_\Omega|\frac{du}{dx}|$ I have ...
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Holder continuity of $\frac{x}{|x|^3} \ast f$ with $f \in C^1_0$ in $\mathbb{R}^3$

Ok, so I need to show that for $f \in C^1_0(\mathbb{R}^3)$ the convolution with $k(x) := \frac{x}{|x|^3}$ is Holder continuous. The exponent doesn't matter much as long as I can bound it using ...
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Product of Hölder and Sobolev functions

Here $C^{\kappa , \lambda} ( \overline{\Omega} ) = \left\{ h|_{\overline{\Omega}} :h \in C^{\kappa , \lambda} ( \mathbb{R}^{n} ) \text{ and } h \text{ has compact support} \right\}$ denotes $\kappa$ ...
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Compact, continuous embeddings of $H^s := W^{s,2} \leftrightarrow C^{(\alpha)}$

The sobolev-space $H^s([-\pi,\pi])$ can be embedded into $C^{(\alpha)}([-\pi,\pi])$ (space of $\alpha$-Hölder-continuous functions) and vice-versa. My question is for which exponents $s, \alpha$ can ...
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Hölder Continuity between metric spaces

Hölder continuity has never appeared in my formal education and the wikipedia article seems insufficiently general. I want to make sure that this definition of Hölder continuity is correct and ...
Convergence of sequence in $C^{k, \alpha}$ composed with $C^\infty$ function
Is it true that if a sequence $u_n \to u$ in $C^{k, \alpha}$ norm, and if you have a function $f \in C^\infty$, then $f(u_n) \to f(u)$ in $C^{k, \alpha}$? I think so, since this is true for ordinary ...