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I'm trying to show that the function $f(x) = x \ln(x)$ is Hölder continuous on $(0,1) $ for $0 < \alpha < 1$.

I must be missing something, because I don't really understand how the definition makes sense. For a function to be Hölder continuous, $ (\forall x,y \in (0,1))\bigl(|f(x) - f(y)| \leq C|x-y|^\alpha\bigr) $.

Then if I divide both sides by $|x-y|^\alpha$, $$\frac{|f(x)-f(y)|}{|x-y|^\alpha} \leq C.$$ But as I can make $|x-y|$ arbitrarily small, how does this make the fraction bounded? For this definition to make any sense does this mean that $|f(x) - f(y)|$ must tend to $0$ faster than $|x-y|^\alpha$?

In regards to my original problem, clearly $|f(x) - f(y)| \leq \frac{1}{e}$. So how do I show itis Hölder continuous? Does showing that $$\lim_{x \to y} \frac{|f(x)-f(y)|}{|x-y|^\alpha} = 0$$ suffice along with the boundedness, or am I on the completely wrong track?

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1 Answer 1

Yes, when $x-y$ is small, the definition dictates that $f(x)-f(y)$ should be small in a certain way. This is what the definition of uniform continuity does as well, with its $\epsilon$ and $\delta$.

Maybe you should start with something simple and concrete: show that the function $f(x)=x^2$ on the interval $(0,1)$ satisfies $|f(x)-f(y)|\le 2|x-y|$.

Then proceed to something harder: show that $f(x)=\sqrt{x}$ is Hölder continuous with $\alpha=1/2$.

As for your original question, the best way to handle it is with Hölder's inequality. Road map:

  • let $p=1/\alpha$
  • observe that $|f'|^{p}$ is an integrable function on $(0,1)$
  • estimate $|f(x)-f(y)|$ by the integral of $|f'|$ over the interval between $x$ and $y$
  • apply Hölder's inequality to $|f'|\cdot 1$, raising $|f'|$ to power $p$. This will yield a factor of $|x-y|^{1/p}$ as required.
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