Let $p \in [1,\infty)$. Does there exist $C>0$ such that for every $f \in W^{1,p}([0,1],\mathbb{R})$ we have $$\|f\|_{L^\infty}\leq C\|f\|_{L^p}^{1-\frac{1}{p}}\|f\|_{W^{1,p}}^{\frac{1}{p}}?$$ My Motivation is the following: For $m \in \mathbb{N}$ I have an approximation operator $Q^h$ depending on the meshwidth $h>0$ such that for every $l\leq m$ there exist $C>0$ such that for every $f \in W^{m,p}$ we have $$\|f-Q^hf\|_{W^{l,p}}\leq Ch^{m-l}\|f\|_{W^{m,p}}.$$ Now I want to show that $\|f-Q^hf\|_{W^{m-1,\infty}} \rightarrow 0$. The statement above would imply that $$\|f-Q^hf\|_{W^{m-1,\infty}}\leq C h^{1-\frac{1}{p}}\|f\|_{W^{m,p}}.$$
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$\begingroup$ Roughly, yes, but your exponents are wrong. You can always figure out what the exponents should be by considering how both sides scale when you rescale $f$ or space. The exponents do have to add up to $1$ as yours do. But when you consider what happens when you rescale space, you don't get $(p-1)/p$ and $1/p$ but $1-a$ and $a$ for another value of $a$. $\endgroup$– Deane YangCommented Apr 16, 2014 at 21:26
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$\begingroup$ Also, the inequality does not hold for all $p \ge 1$. $p$ must be greater than $n$. You can see this by testing the inequality with $f = r^{-\alpha}$. $\endgroup$– Deane YangCommented Apr 16, 2014 at 21:28
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$\begingroup$ @Deane Yang: But $n=1$ year, isn't it? $\endgroup$– Delio MugnoloCommented Apr 16, 2014 at 23:08
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$\begingroup$ Delio, yes, you're right. I didn't notice the dimension. $\endgroup$– Deane YangCommented Apr 17, 2014 at 2:54
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$\begingroup$ There is a very easy way to get this kind of interpolation inequality. Under these assumptions, the function is Holder continuous (with constant depending on the $W^{1,p}$ seminorm). If its maximum is 1, then you can put a Holder modulus underneath the graph of the function, centered at the point where the max occurs. When you see how much area lies under the graph of the (pth power of) this "tent", you get a lower bound for the L^p norm of the function. I believe you get something close to the conjectures inequality in the question, but I am too lazy to check how the exponents work out. $\endgroup$– Scott ArmstrongCommented Apr 17, 2014 at 3:36
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Yes. This is a known inequality, called e.g. "Gagliardo-Nirenberg inequality" in Brezis' book, Comment 1 to Chapter 8, page 233.
