10
$\begingroup$

Assume $0 < \beta < \gamma \le 1$. Prove the interpolation inequality $$\|u\|_{C^{0,\gamma}(U)} \le \|u\|_{C^{0,\beta}(U)}^{\frac{1-\gamma}{1-\beta}} \|u\|_{C^{0,1}(U)}^\frac{\gamma-\beta}{1-\beta}.$$

From PDE Evans, 2nd edition: Chapter 5, Exercise 2

I would love to employ Hölder's inequality which can easily justify this inequality. But Hölder's inequality requires $u \in L^p(U), v \in L^q (U)$. Instead, this problem has $u \in C^{0,\beta}(U) \cap C^{0,1}(U)$.

The textbook gives the definition of \begin{align} \|u\|_{C^{0,\gamma}(\bar{U})} &:= \|u\|_{C(\bar{U})}+[u]_{C^{0,\gamma}(\bar{U})} \\ &= \sup_{x \in U} |u(x)|+\sup_{\substack{x,y \in U \\ x \not= y}} \left\{\frac{|u(x)-u(y)|}{|x-y|^\gamma} \right\} \end{align}

All I know so far is that, given $0 < \beta < \gamma \le 1$ ...

  • If $|x-y|<1$, then $\frac 1{|x-y|^\gamma}<\frac 1{|x-y|^\beta}$, which means $\|u\|_{C^{0,\gamma}(U)} < \|u\|_{C^{0,\beta}(U)}$.
  • If $|x-y|\ge 1$, then $\frac 1{|x-y|^\gamma}\le \frac 1{|x-y|}$, which means $\|u\|_{C^{0,\gamma}(U)} \le \|u\|_{C^{0,1}(U)}$.
$\endgroup$
2
  • $\begingroup$ Are you sure that your last "which means" is valid? $\endgroup$
    – Pedro
    Sep 2, 2015 at 3:20
  • $\begingroup$ @Pedro Please ignore my work. It was wrong. $\endgroup$
    – Cookie
    Sep 2, 2015 at 6:50

2 Answers 2

12
$\begingroup$

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 $$ \| u\|_\gamma \leq |u|_\infty^{1-t}|u|_\infty^t + [u]_{\beta}^{1-t}[u]_{1}^t =:a_1^{1-t}b_1^t+a_2^{1-t}b_2^t. $$ Now just write $A=a_1+a_2$, $A_i=a_i/A$, then the RHS becomes $$ A^{1-t}\left( A_1\left( \frac{b_1}{A_1}\right)^t + A_2 \left( \frac{b_2}{A_2}\right)^t \right) \leq A^{1-t}(b_1+b_2)^t, $$ where the last inequality is the concavity of the function $s\mapsto s^t$, since $A_1+A_2=1$. Combining all this gives the desired inequality.

$\endgroup$
3
  • $\begingroup$ I followed through your steps and got the desired inequality. I do have two questions though: 1) How do we know that the mapping $s \mapsto s^t$ is concave and not convex? I thought that $s^t$ was exponential. 2) Given that it's concave as you say, how do I relate the concavity equation $tf(x)+(1-t)f(y) \le f(tx+(1-t)y)$ to what you wrote: $$A^{1-t}\left( A_1\left( \frac{b_1}{A_1}\right)^t + A_2 \left( \frac{b_2}{A_2}\right)^t \right) \leq A^{1-t}(b_1+b_2)^t$$ $\endgroup$
    – Cookie
    Jan 11, 2015 at 5:30
  • 2
    $\begingroup$ 1) The mapping $s\mapsto s^t$ is not exponential, you're thinking of $t\mapsto s^t$. To see concavity we can calculate the second derivative on $(0,\infty)$, which is $t(t-1)s^{t-2}<0$. 2) To relate $sf(x)+(1-s)f(y)\leq f(sx+(1-s)y)$ to the present situation, consider $s=A_1$, $1-s=A_2$, $x=b_1/A_1$, $y=b_2/A_2$ and $f(x)=x^t$. $\endgroup$
    – Jose27
    Jan 11, 2015 at 5:38
  • $\begingroup$ I think the proof applies to the case $\beta = 0$ with the difference that the main inequality gets a constant different from one in front of the product of norms on the right hand side. $\endgroup$
    – Ceka
    Mar 19, 2020 at 16:01
0
$\begingroup$

Here is another approach for solution. First notice the inequality seems to imply that $$f(\gamma) := ||u||_{C^{0,\gamma}(\overline{U})}$$ is a logarithmically convex function. If a function is twice differentiable, then it is logarithmically convex on intervali $I\subset \mathbb{R}$ if and only if $$ f''f\geq (f')^2 $$ Also a suppremum of a set of log-convex functions is a log-convex (wiki). So to deduce if $f$ is log-convex, it is enough to investigate Hölder seminorm $[u]_{C^{0,\gamma}(\overline{U})} =\sup_{x\neq y}\big( \frac{|u(x)-u(y)|}{|x-y|^\gamma}\big )$ only. $$g(\gamma) := \frac{|u(x)-u(y)|}{|x-y|^\gamma}$$ Notice that a function $g$ is twice differentiable with respect to $\gamma$.

\begin{align*} \frac{d}{d\gamma} g &= -\log(|x-y|)\frac{|u(x)-u(y)|}{|x-y|^{\gamma}}\\ \frac{d^2}{d\gamma^2} g &= \log^2(|x-y|)\frac{|u(x)-u(y)|}{|x-y|^{\gamma}}\\ \end{align*} Now to test log-convexity of $g$, input it to the inequality above. \begin{align*} g''g \geq (g')^2 \iff 1\geq 1 \end{align*} This implies that Hölder seminormi is a log-convex function with respect to $\gamma$, because the supremum is also a log-convex. Now the Hölder norm $$||u||_{C^{0,\gamma}(\overline{U})} = ||u||_\infty + [u]_{C^{0,\gamma}(\overline{U})}$$ is a log-convex too, because adding a constant $||u||_\infty$ doesn't change it. All the heavy lifting is done, and from the logarithmic convexity of the norm the inequality follows. \begin{align*} \log\bigg ( f\big( (1-t)\beta + t \big) \bigg ) &\leq (1-t)\log\big( f(\beta) \big ) + t \log \big (f (1) \big )\\ &\implies\\ f\big( (1-t)\beta + t \big) &\leq f(\beta)^{1-t}f(1)^t \end{align*} Because the assumption $0<\beta\leq \gamma \leq 1$, there exists $t_\gamma$ such that $$(1-t_\gamma)\beta + t_\gamma = \gamma$$ Input this to the earlier inequality to get the interpolation inequality. \begin{align} f(\gamma) = ||u||_{C^{0,\gamma}(\overline{U})} \leq ||u||_{C^{0,\beta}(\overline{U})}^\frac{1-\gamma}{1-\beta}\, ||u||_{C^{0,1}(\overline{U})}^\frac{\gamma-\beta}{1-\beta} = f(\beta)^\frac{1-\gamma}{1-\beta} \, f(1)^\frac{\gamma-\beta}{1-\beta} \end{align}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .