22

Or is there a simpler way to do it (...)? This was asked on the site before and requires no differentiability argument nor the mean value theorem, nor anything but the hypothesis, really... For every $x\ne y$, split the interval $[x,y]$ into $n$ subintervals of length $\frac1n\cdot|x-y|$. The Hölder condition on each interval yields a bound $K\cdot \left(\...


19

Hint: why is it not possible to find a $C$ such that $$ |\sqrt{x} - \sqrt{0}|\leq C|x-0| $$ For all $x \geq 0$? As a general rule: Note that a differentiable function will necessarily be Lipschitz on any interval on which its derivative is bounded. In response to the wikipedia excerpt: "This function becomes infinitely steep as $x$ approaches $0$" is ...


17

We have to show the following: given a Cauchy sequence $(u_n)_{n\in \mathbb N}$ in $C^{k,\gamma}(\overline U),$ there is a $u \in C^{k,\gamma}(\overline U)$ such that $\lim_{n\rightarrow\infty}u_n = u$ in $C^{k,\gamma}(\overline U).$ Note that the Hölder norm is the "sum" of the $C^k$ norm (i.e. sup-norm up to the $k$-th derivatives) and the Hölder ...


13

If $F \subset [a,b]$ is an $F_{\sigma}$-set then $F = \bigcup_{n=1}^{\infty} K_n$ with $K_n$ compact (since closed subsets of $[a,b]$ are compact). By continuity of $f$ the set $f(F) = \bigcup_{n=1}^{\infty} f(K_n)$ is the countable union of the compact sets $f(K_n)$ and hence $f(F)$ an $F_\sigma$-set, so $f(F)$ is measurable. If $N \subset [a,b]$ is a null ...


13

You have $${\sqrt{1/n} - \sqrt{0}\over{1/n - 0}} = {1/\sqrt{n}\over {1\over n}} = \sqrt{n}.$$ This ratio can be made as large as you like by choosing $n$ large. Therefore the square-root function fails to be Lipschitz.


10

As mentioned in the wiki entry, the function defined by $f(x)=\cases{{1\over \ln x},&$0<x\le 1/2$\cr \strut0, &$x=0$}$ is an example of a uniformly continuous function that is not Hölder continuous for any $\alpha>0$. $f$ is continuous on $[0,1/2]$; and thus, since $[0,1/2]$ is compact, uniformly continuous on $[0,1/2]$. But, $f$ is ...


10

Here is an example: $N=1$, $E=[0,e^{-3})$ and $$f(x)=\left\{\begin{array}{ccc}(\log x)^{-2}&,& x\in(0,e^{-3})\\ 0&,&x=0\end{array}\right..$$ For every $\alpha>0$, $$\lim_{x\to 0^+}\frac{|f(x)-f(0)|}{x^\alpha}=\infty,$$ so $f$ is not $\alpha$-Hölder continuous. Now let us show that $f$ is Dini continuous. To begin with, note that $$f'(x)...


9

HINT: If I assume you already have proved or known that Denote $C^k(\Omega)$ as the space for functions bounded and continuous up to $k$-th derivative, $C^k(\Omega)$ equipped with the $\sum\limits_{|\alpha|\leq k} \sup\limits_{x\in \Omega} |\partial_{\alpha}(\cdot)|$ is a Banach space. Then you need to check if the following facts still hold after the ...


9

This is a theorem of Bernstein, found, for example, in Katznelson's Introduction to Harmonic Analysis. A simple-minded approach would be to shift the variable in $$c_n=\frac{1}{2\pi}\int_0^{2\pi} f(x)e^{-inx}\,dx\tag1$$ to $y=x+\pi/n$, thus obtaining $$c_n=\frac{1}{2\pi}\int_{0}^{2\pi} f_{\pi/n}(y)e^{-iny+\pi i}\,dy\tag2$$ where I write $f_h(x)=f(x-h)$. ...


9

Notice that $$ f'(x)=\begin{cases} 2x\sin\frac1x-\cos\frac1x &\text{ if } x\ne0\\ 0 &\text{ if } x=0 \end{cases}. $$ Therefore $$ |f'(x)|\le 2|x|\cdot\left|\sin\frac1x\right|+\left|\cos\frac1x\right|\le 3 \quad \forall x\ne 0. $$ Hence, $$ |f'(x)|\le 3 \quad \forall x \in \mathbb{R}. $$ It follows that $$ |f(x)-f(y)|\le \sup_{\min\{x,y\} \le z\le \...


8

Your answer depends, as you guessed on the process $u \mapsto \sigma_u$. You can amplify your approach using the Ito-isometry with the BDG-inequality: \begin{align} \mathbb{E}[|\int_0^t\sigma_udW_u - & \int_0^s\sigma_u \,dW_u|^{2p}] \stackrel{\text{BDG}}{\leq} c(p) \mathbb{E}[(\int_s^t |\sigma_u|^2 \, du)^p] \\ & \leq c(p) \mathbb{E}[(t-s)^{p-1} \...


8

I don't know how much this will help, but: The simplest instance of a Holder norm controlled by a Sobolev norm is that of the Fundamental Theorem of Calculus. Let us suppose we have a function $f:\mathbb{R}\to\mathbb{R}$ such that its distributional derivative $f'$ is defined, and such that we have $\int |f'|^p \mathrm{d}x < \infty$. Then we can estimate ...


8

Suppose that $\sqrt{x}$ is a Lipschitz function, then there exists $C$ such that $$\Big|\frac{\sqrt{y}-\sqrt{x}}{y-x}\Big| \le C$$ Now, Let $y=2x$, so $$(\sqrt{2}-1)x^{-\frac{1}{2}}\le C$$ Letting $x→0$ gives a contradiction.


8

Set $t\in (0,1)$ such that $(1-t)\beta + t = \gamma$, then we have $$ \frac{|u(x)-u(y)|}{|x-y|^\gamma}= \left( \frac{|u(x)-u(y)|}{|x-y|^{\beta}} \right)^{1-t}\left( \frac{|u(x)-u(y)|}{|x-y|}\right)^t\leq [u]_{\beta}^{1-t}[u]_{1}^t. $$ Also, we always have $| u|=|u|^{1-t}|u|^t$. Combining this with the previous estimate we get $$ \| u\|_\gamma \leq |u|_\infty^...


8

For $a\in(0,1)$, define $u_{a}(t)=0$ for $t\in[0,a]$ and $u_{a}(t)=(t-a)^{\alpha}$ for $t\in(a,1]$, then for $0<a<b<1$, we have \begin{align*} \|u_{a}-u_{b}\|&=\sup_{0\leq x_{1}\ne x_{2}\leq 1}\dfrac{|(u_{a}-u_{b})(x_{1})-(u_{a}-u_{b})(x_{2})|}{|x_{1}-x_{2}|^{\alpha}}\\ &\geq\dfrac{|(u_{a}-u_{b})(b)-(u_{a}-u_{b})(a)|}{|b-a|^{\alpha}}\\ &...


7

I think you should have a look at Sergiu Klainerman's 30-page article in the Princeton Companion to Mathematics. It's a recent and well-written overview-like treatment of PDEs. There is also an extended version of the article to be found on Klainerman's website.


7

Let $f: [a,b] \rightarrow \mathbb{R}$ be a $C^1$-function, i.e., $f'$ exists at every point and is continuous. Since $f': [a,b] \rightarrow \mathbb{R}$ is continuous, it is bounded: there is a constant $M$ such that $|f'(x)| \leq M$ for all $x \in [a,b]$. By the Mean Value Theorem, for any $a \leq x \leq y \leq b$, $|f(y)-f(x)| = |f'(c)||y-x| \leq M|y-x|...


7

Assume $x>y$. $$x^s-y^s \leq C(x-y)^s$$ $$\Leftrightarrow x^s \leq C(x-y)^s+y^s$$ Claim: this holds for all $(x,y)$ when $C=1$. Proof: Because $0<s\leq 1$, for all $a,b \geq 0$ $$(\frac{a}{a+b})^s+(\frac{b}{a+b})^s \geq 1$$ $$\Leftrightarrow a^s+b^s \geq (a+b)^s$$. Now set $a=x-y$ and $b=y$.


6

Presumably $\Delta(x,y) = \sup_{a \leq t \leq b} |x(t) - y(t)|$. For a): as Giuseppe said: use the Arzelà-Ascoli theorem (for $f \in K$ you have $L \leq 1$). For b): For $p \in [a,b]$ consider the function $f_p(t) = |p-t|$ and show that for $p \neq q$ $$ d_{0,1}(f_p,f_q) = \underbrace{\sup_{t} \cdots}_{= d(p,q)} + \underbrace{\sup_{s,t} \cdots}_{\geq 2} \...


6

A sufficient condition is: the sequence $f_n$ is bounded in $C^{0,\gamma+\epsilon}$ for some $\epsilon>0$. For a proof, see Is there a reference for compact imbedding of Hölder space? A slightly weaker sufficient condition: there exists a modulus of continuity $\omega:(0,\infty)\to(0,\infty)$ such that $\omega(\delta)=o(\delta^\gamma) $ as $\...


6

You can find a counterexample in this link, however I would like to note that for every $0<\beta<\alpha$ we do have convergence. Indeed, define for $x\neq y$ $$T(\epsilon,x,y)=\int_{B_1}\left|\varphi(z)\frac{g(x-\epsilon z)-g(x)-(g(y-\epsilon z)-g(y))}{|x-y|^\beta}\right|.$$ Case 1: $|x-y|\ge \delta_1$ In this case we have that for any $\eta>0$, ...


6

Yes, if you omit the middle term you get an equivalent norm. (The reason the middle term is typically included is just to make various things simpler.) This is a generalization of the classical Landau inequality, which says that a bound on $f$ and a bound on $f''$ imply a bound a $f'$. One can give a simple proof like so: Assume $f'$ is "large" at a point. ...


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