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 $\|u\|_{C^{0,\gamma}(U)} \le \|u\|_{C^{0,1}(U)}$.
  • $\begingroup$ Are you sure that your last "which means" is valid? $\endgroup$ – Pedro Sep 2 '15 at 3:20
  • $\begingroup$ @Pedro Please ignore my work. It was wrong. $\endgroup$ – Cookie Sep 2 '15 at 6:50

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.

  • $\begingroup$ 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 '15 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 '15 at 5:38

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