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Consider the function $f:[0,1]\rightarrow \mathbb{R}$ given by

$$f(x)= x\log(x).$$

This function is not Lipschitz continuous at zero, but apparently Hölder continuous for any $\alpha<1$, i.e. there is a constant $C_{\alpha}$ such that

$$\vert f(x)-f(y)\vert \le C_{\alpha} \vert x-y \vert^{\alpha},$$

but how can one show the Hölder continuity?

Please let me know if you have any questions or remarks.

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  • $\begingroup$ @absolute0 sure it is continuous at zero, in fact the function is zero at zero. $\endgroup$
    – Sascha
    Mar 11 at 4:57
  • $\begingroup$ Yes, yes my bad! I'll retract that. $\endgroup$
    – absolute0
    Mar 11 at 5:37
  • $\begingroup$ @absolute0 Also notice that If a function is Holder continuous on a bounded set $A$, then it is uniformly continuous on $A$.. $\endgroup$
    – Medo
    Mar 11 at 9:14
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Same plan of attack as $x^\alpha$ as an example of an $\alpha$-Hölder continuous function, which should be reminiscent of proofs using e.g. mean value theorem and the sort.

Note $-x\log x\ge 0$ for $x\in[0,1]$ (with the obvious redefinition at zero). Let $0\le x_0<x\le 1$ and consider $g:[x_0,1]\to\mathbb R$ defined by $$g(x) = -x\log x + x_0 \log x_0 + (x-x_0)\log(x-x_0).$$ Then $$g'(x) = (-\log x - 1) - (-\log(x-x_0) - 1) = \log (x-x_0) - \log x\le 0,$$ which follows because of the monotonicity of $\log$ and $0<x-x_0\le x $. Thus $g$ is decreasing, with $g(x_0)=0$. This implies for any $0\le x<y\le e^{-1}$, (note $-t\log t$ is increasing on $[0,e^{-1}]$)

$$ \frac{(-y\log y) - (-x\log x)}{-(y-x)\log(y-x)} = \frac{|y\log y - x\log x|}{|y-x|\lvert\log \lvert y-x \rvert \rvert} \le 1.$$ Therefore, $$ \sup_{\substack{y\neq x\\x,y\in[0,e^{-1}]}} \frac{|y\log y - x\log x|}{|y-x|\lvert\log \lvert y-x \rvert \rvert} \le 1.$$ For points away from the origin, it is easy to check that the function is Lipschitz (in fact smooth). Therefore, it follows after some routine case checking (see end of answer) that for some $C>0$, $$ \sup_{\substack{y\neq x\\x,y\in[0,1]}} \frac{|y\log y - x\log x|}{|y-x|\max(\lvert\log \lvert y-x \rvert \rvert,1)} \le C.$$ (The maximum is just to avoid 'irrelevant' considerations when $x$ and $y$ are far apart i.e. $|x-y|\to 1$. A function satisfying this type of inequality is said to be 'Log-Lipschitz'.)

This is in fact stronger than being $\alpha$-Hölder continuous: since $ |r|\max(|\log |r||,1)\le C_\alpha |r|^\alpha$ for any $0<\alpha<1$, we immediately have $$ \sup_{y\neq x} \frac{|y\log y - x\log x|}{|y-x|^\alpha} \le C_\alpha \sup_{y\neq x} \frac{|y\log y - x\log x|}{|y-x|\lvert\log \lvert y-x \rvert \rvert} \le CC_\alpha. $$ Indeed, setting $r=e^{-t}$ for $t> 0$, the above is the statement $\max(t,1)e^{-t} \le C_\alpha e^{-\alpha t} \iff \max(t,1) \le C_\alpha e^{(1-\alpha)t}$; so the statement holds with $C_\alpha = \frac1{1-\alpha}$.


(Routine case checks) for any $0<\epsilon<e^{-1}$ the function $x\log x$ is Lipschitz on $[\epsilon,1]$ with constant $\sup_{x\in [\epsilon,1]}| \log x +1|=\max(1,-\log\epsilon-1)=:C_1.$ (This is because the function has Lipschitz constant 1 on $[e^{-1},1]$.) Since we already have control for the region $0<x<y<e^{-1}$, this gives the required control for all $0\le x<y\le 1$ with $|x-y|<e^{-1}-\epsilon=:C_2.$ For the remaining regions of $(x,y)\in[0,1]^2$, it suffices to obtain control in the region $|x-y|>C_2$ where we can crudely bound with $$\frac{|x\log x-y\log y|}{|x-y|\max(\lvert\log\lvert x-y\rvert\rvert,1)}\le \frac{\max_{x\in[0,1]}|x\log x|-\min_{x\in[0,1]}|x\log x|}{|x-y| }=\frac1{eC_2}.$$ Thus, for some fixed sufficiently small $\epsilon$, we can take $$C=\max(1,C_1,\frac1{eC_2})=\max(1,-\log\epsilon-1,\frac1{1-e\epsilon}).$$

Setting $\epsilon=e^{-2}$ gives $C= 2/(1-e^{-1})\approx 1.58$. Numerically, the optimal choice of $\epsilon$ gives $C\approx 1.35$. but $C=1$ seems to be true even for the whole interval $[0,1]$.

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    $\begingroup$ Actually, I am not entirely sure. Set $x=1$, then your supremum is $$\sup_y \frac{y \vert \log(y)\vert}{\vert y-1\vert \vert \log(y-1)\vert},$$ then when $y$ goes to zero, the nominator goes to zero like $y \vert \log(y)\vert$ whereas by Taylor expansion the denominator goes to zero like $\frac{1}{y},$ this looks rather singular $\endgroup$
    – Sascha
    Mar 11 at 14:11
  • $\begingroup$ Hmm. Ok so then the right way is to do the above only for small x and y then use smoothness away from 0. I’ll fix it later need to sleep! $\endgroup$ Mar 11 at 14:41
  • $\begingroup$ @Sascha I've fixed it. Generally when you use expressions like $\frac{|f(x)-f(y)|}{\omega(|x-y|)}$ to define spaces, you want either to restrict $x,y$ to be sufficiently close (because we are interested in the local regularity at each point), or otherwise make the denominator cause no trouble when $|x-y|$ is large. (Also a rather annoying typo- the argument of the logarithms need absolute values as well) $\endgroup$ Mar 12 at 2:56
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This can be seen by looking at the derivative: $$ f'(x) = \log x + 1, $$ which is controlled by the derivative of $f_\alpha(x) = x^\alpha$ (for any $0 < \alpha < 1$), which is $f_\alpha'(x) = \alpha x^{\alpha-1}$. Indeed, you should be able to check that $\left| \frac{f'(x)}{f_\alpha'(x)} \right|$ is bounded by some constant $C_\alpha > 0$ independent of $x \in (0,1)$ (but depending on $\alpha$).

Knowing that $f_\alpha$ is $\alpha$-Holder continuous, you can infer the same about $f$: $$ |f(x)-f(y)| \le \int_x^y |f'(t)| dt \le C_\alpha \int_x^y |f_\alpha'(t)| dt = C_\alpha |f_\alpha(x)-f_\alpha(y)| \le C_\alpha C'_\alpha |x-y|^\alpha, $$ of course also using monotonicity of $f_\alpha$.

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