Determining if $\|f\|_\infty \leq C\, \|f\|_{2}^{2/3} $ holds under $f(0) = f(1) = 0$, $\|f'\|_2 \leq 1$

Question

Suppose $f \colon [0, 1] \to \mathbb{R}$ is continuously differentiable, and satisfies $f(0) = f(1) = 0$ and $\|f'\|_2 \leq 1$.

I am wondering if it there is a constant $C > 0$ such that for all such $f$, we have $$ \sup_{x \in [0, 1]} |f(x)| \equiv \|f\|_\infty \leq C\, \|f\|_{2}^{2/3}. $$ (Here $\|g\|_2^2 = \int_0^1 g^2$. )

This holds if $f$ is additionally Lipschitz as is shown in this post.

This seems superficially related to the Gagliardo-Nirnberg/Sobolev Interpolation inequalities, but I wasn't able to get this from either of those directly.

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EDIT: It seems this cannot hold, but I would still like to understand why this doesn't hold, but it does hold under Lipschitz condition.

Here is my reasoning as to why it cannot hold. Suppose it does. Then by defining $g = f/\|f'\|_2$, we have $\|g'\|_2 = 1$, so the inequality implies that for any $f$ continuously differentiable with $f(0) = f(1) = 0$ that $$ \|f\|_\infty \leq C \, \|f\|_2^{2/3} \|f'\|_2^{1/3}. $$ However, from this inequality, we can now consider $f_\lambda(x) = f(\lambda x)$ for $x \in [0, 1/\lambda]$ and $0$ else, for $\lambda > 1$. Then $$ \|f\|_\infty = \|f_\lambda\|_\infty \leq C \, \|f_\lambda\|_2^{2/3} \|f_\lambda'\|_2^{1/3} = C\, \|f\|_2^{2/3} \|f'\|_2^{1/3} \lambda^{-1/6}. $$ Taking $\lambda \to \infty$ shows $f \equiv 0$, which is obviously absurd.

Answer 1

Counterexample: a smoothed version of the function $f$ given by the formula $$f(x)=(h/2)^{1/2}(1-\max(0,|x/h-1|))\tag{1}\label{1}$$ with $h\downarrow0$.

Indeed, if we had $$\|f\|_\infty\le C\,\|f\|_{2}^{2/3} \tag{2}\label{2}$$ for some real $C>0$ and all continuously differentiable functions $f\colon[0,1]\to\Bbb R$ such that $f(0)=f(1)=0$ and $\|f'\|_2\le1$, then, by approximation, we would have \eqref{2} for $f$ as in \eqref{1}, because for such $f$ we have $f(0)=f(1)=0$ and $\|f'\|_2=1$. However, for such $f$, we also have $\|f\|_\infty=(h/2)^{1/2}$, whereas $\|f\|_{2}^{2/3}\asymp h^{2/3}=o(h^{1/2})$ as $h\downarrow0$.

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This shows that the best (largest) exponent $p>0$ for which $$\|f\|_\infty\le C\,\|f\|_{2}^p \tag{2a}\label{2a}$$ can hold (for some real $C>0$ and all continuously differentiable functions $f\colon[0,1]\to\Bbb R$ such that $f(0)=f(1)=0$ and $\|f'\|_2\le1$) is $\le1/2$.

On the other hand, for any continuously differentiable function $f\colon[0,1]\to\Bbb R$ such that $f(0)=f(1)=0$ and $\|f'\|_2\le1$ and any $x\in[0,1]$, $$f(x)^2=\int_0^x dt\,\frac{d}{dt}\,f(t)^2 =2\int_0^x dt\,f(t)f'(t)\le2\|f\|_2\|f'\|_2 \le2\|f\|_2,$$ so that \eqref{2a} holds with $p=1/2$ and $C=\sqrt2$. Thus, the best $p$ in \eqref{2a} is $1/2$.

Answer 2

Let me try to give an intuitive explanation for why it is conceivable that inequality like this can hold for Lipschitz functions (with bounded Lipschitz norm) and yet fail for $f$ with $f'\in L^2$. The idea is really simple: the Morrey's inequality (Sobolev's inequality for $p > n$).

In our situation it says that if $f'\in L^2$ then the function $f$ is $\frac{1}{2}$-Hölder. Moreover, it is known that Sobolev's and Morrey's inequalities are essentially sharp (in our case, we can not have a uniform estimate with a better modulus of continuity, say). So informally we are asking if there're inequalities which work for Lipschitz functions and fail for $\frac{1}{2}$-Hölder, and for me it seems natural that it is a real possibility, since the Lipschitz condition is strictly stronger.

Alternatively, you can go in the other direction and say that $f$ is Lipschitz if and only $f'\in L^\infty$ and now we are asking ourselves, why some estimates might work for $f' = g\in L^\infty$ and fail for general $g\in L^2$, which is now even more clear intuitively by duality.