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Let $I\subset \mathbb R$ be an interval, $1\leq p\leq \infty$, and $W^{1,p}(I)$ the usual Sobolev space. It is known that the injection $W^{1,p}(I)\hookrightarrow L^\infty(I)$ holds, i.e. there exists $C$, depending on $I$, such that for all $u\in W^{1,p}(I)$ we have $$ \|u\|_{L^\infty(I)}\leq C\|u\|_{W^{1,p}(I)}. $$ What is the optimal $C$ (in terms of $I$) and what are the optimizers for the above inequality ?

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  • $\begingroup$ For $p = 1$ the optimal estimate is fundamental theorem of calculus, and there is no optimizer (the optimizer would be the heaviside distribution centered "at the right end point" of your interval $I$, which is not in $W^{1,1}$). For $p > 1$ the oscillation estimate $$|u(b) - u(a)| \leq |b-a|^{1 - 1/p} \left(\int_a^b |u'|^p\right)^{1/p}$$ is sharp. $\endgroup$ Commented Jan 18, 2017 at 15:57
  • $\begingroup$ On the other hand, we have the mean estimate $$ \frac{1}{|b-a|} \int_a^b u \leq \frac{1}{|b-a|^{1/p}} \|u\|_{L^p} $$ which implies that $$ \|u\|_\infty \leq \max(|b-a|^{1-1/p}, |b-a|^{-1/p}) \|u\|_{W^{1,p}} $$ $\endgroup$ Commented Jan 18, 2017 at 16:28
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    $\begingroup$ For $|b-a| < 1$ this implies that the optimal constant is $|b-a|^{-1/p}$ and the optimizer is the constant function. But this is an artifact of the fact that your initial inequality has the wrong scaling: for example, the following slightly sharper inequality holds: $$\|u\|_{L^\infty}^p \leq C \|u\|_{L^p}^{p-1} \|u\|_{W^{1,p}} $$ So the optimizer for your inequality is going to depend somewhat on the quirkiness of how scaling interacts with the definition of your norm (how does the $\|u\|_{L^p}$ and $\|u'\|_{L^p}$ portion contribute to the computation of the norm) and will not be geometric. $\endgroup$ Commented Jan 18, 2017 at 16:37
  • $\begingroup$ Dear @WillieWong, thanks for your answer and for pointing out that the minimizer will depend on the definition of the norm. I thought at first that the answer to this question should be well-known, but apparently not. $\endgroup$
    – JKLM
    Commented Jan 20, 2017 at 15:35

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