1
$\begingroup$

This is a segment of a proof to a theorem from PDE Evans, 2nd edition, page 156:

THEOREM 5 (Asymptotics in $L^\infty$ norm). There exists a constant $C$ such that $$|U(x,t)| \le \frac{C}{t^{\frac 12}}$$ for all $x \in \mathbb{R}, t>0$.

Proof. Set $$\sigma:=F'(0); \tag{58}$$ then $$G(\sigma)=(F')^{-1}(\sigma)=0,\tag{59}$$ and therefore $$L(\sigma)=\sigma G(\sigma)-F(G(\sigma))=0, L'(\sigma)=0.\tag{60}$$

In view of $(60)$ and the uniform convexity of $L$, \begin{align} tL (\frac{x-y}{t} ) &= tL \left(\frac{x-y-\sigma t}{t}+\sigma \right) \\ &\ge t \left[\require{cancel} \cancelto{0}{L(\sigma)}+\cancelto{0}{L'(\sigma)}\left(\frac{x-y-\sigma t}{t} \right) + \theta\left(\frac{x-y-\sigma t}{t} \right)^2 \right]\\ &= \theta \frac{|x-y-\sigma t|^2}{t} \end{align}

The proof continues on in the text but I stopped here as my question is regarding the second-to-last step shown above here.

I strongly think that the uniform convexity of $L$ is used to assert this inequality: $$tL \left(\frac{x-y-\sigma t}{t}+\sigma \right) \ge t \left[ \require{cancel} \cancelto{0}{L(\sigma)}+\cancelto{0}{L'(\sigma)}\left(\frac{x-y-\sigma t}{t} \right) + \theta\left(\frac{x-y-\sigma t}{t} \right)^2 \right]$$ But my question is how so exactly?

$\endgroup$
  • $\begingroup$ By applying Taylor's formula to the expansion of $L$ around the minimum at $\sigma$? $\endgroup$ – Harald Hanche-Olsen Jun 25 '14 at 20:20
  • $\begingroup$ I'm just thrown off by the $\theta$ ... is it because higher order terms $\ge \theta \left(\frac{x-y-\sigma t}{t} \right)^2$, hence we have that $\ge$ relation in the work process? $\endgroup$ – Cookie Jun 25 '14 at 20:44
  • $\begingroup$ $L(\sigma+\xi)=L(\sigma)+L'(\sigma)\xi+\frac12L'(\sigma+c\xi)\xi^2$ for some $c\in(0,1)$, according to Taylor's formula. There is no need to consider higher order terms, that is all taken care of by this form of the remainder term. $\endgroup$ – Harald Hanche-Olsen Jun 26 '14 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.