Consider $f:[0,1]^d \to \mathbb{R}$. Suppose that $f$ is $L$-Lipschitz w.r.t. the Euclidean norm. Can we provide an upper bound on $\|f\|_\infty$ in terms of $\|f\|_1 := \int_{[0,1]^d} |f(x)|dx$ ?
In dimension 1, I would think that the way to construct such a function $f$ with as large as possible supremum norm, under the constraint that for example $\|f\|_1 = 1$, if $L > 2$, is to pick a $f$ of the form
$$f: x \mapsto \begin{cases} 0 \text{ if } x \in [0, 1-2/L] \\
L(x – 2/L) \text{ if } x \in (1-2/L, 1] \end{cases},$$
which would give us $\|f\|_{\infty} = 2$.
In higher dimensions, I conjecture that the largest supremum norm you can get under $\|f\|_1 = 1$ is achieved by a function whose graph is some type of hyperpyramid, as in the case $d=1$, which would give us that $\|f\|_\infty \lesssim (d/L)^{1/d}$. Would anyone know how to prove this?
This sounds like a result that should be documented, but I couldn't find a good source.
Best Answer
$\newcommand\Om\Omega$Now consider the general case of any natural $d$. Here we will give an upper bound on $\|f\|_\infty$ in terms of $\|f\|_1$, $L$, and $d$. This bound will be optimal up to a factor depending only on $d$; as follows from a comment of yours, such factors do not matter to you. The mentioned bound will be exact in the case $d=1$.
Indeed, let $I:=[0,1]$, $\Om:=I^d$, $$M:=\|f\|_\infty=\max_{x\in\Om}|f(x)|=|f(a)|$$ for some $a=(a_1,\dots,a_d)\in\Om$. Then \begin{equation} |f(x)|\ge h_a(x):=(M-L|x-a|)_+=L(r-|x-a|)_+ \tag{1} \end{equation} for all $x\in\Om$, where $|x-a|$ is the Euclidean norm of $x-a$, $u_+:=\max(0,u)$, and $$r:=M/L$$ (assuming $L>0$). So, $$\frac{\|f\|_1}L=\frac1L\int_\Om|f|\ge\int_\Om dx\,(r-|x-a|)_+ =E(r-R)_+,$$ where $$R:=\sqrt{\sum_1^d(U_i-a_i)^2}$$ and $U_1,\dots,U_d$ are independent random variables each uniformly distributed on $[0,1]$.
Next, $$E(r-R)_+=E\int_0^r dv\,1(R<v)=\int_0^r dv\,P(R<v),$$ \begin{align*} P(R<v)&=P\Big(\sum_1^d(U_i-a_i)^2<v^2\Big) \\ &\ge\prod_1^d P(|U_i-a_i|<v/\sqrt d) \tag{2} \\ &\ge\prod_1^d P(|U_i|<v/\sqrt d) \tag{3} \\ &=\min\Big(1,\frac{v}{\sqrt d}\Big)^d=:Q(v). \end{align*} So, \begin{align*} \frac{\|f\|_1}L&\ge\int_0^r dv\,P(R<v) \\ &\ge\int_0^r dv\,Q(v) \\ &=g(r):=\left\{\begin{aligned} \frac{r^{d+1}}{c_d^{d+1}}&\text{ if }r\le\sqrt d, \\ r-\frac{d\sqrt d}{d+1}&\text{ if }r\ge\sqrt d, \end{aligned} \right. \end{align*} where \begin{equation*} c_d:=((d+1)d^{d/2})^{1/(d+1)}. \end{equation*} Solving now the inequality $\frac{\|f\|_1}L\ge g(r)$ for $r$ and recalling that $r=M/L=\|f\|_\infty/L$, we get \begin{align*} \|f\|_\infty&\le B_0(\|f\|_1,L) \\ &:=Lg^{-1}\Big(\frac{\|f\|_1}L\Big) \\ &=\left\{\begin{aligned} c_d L^{d/(d+1)}\|f\|_1^{1/(d+1)}&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\ %\|f\|_1&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\ \|f\|_1+\frac{d\sqrt d}{d+1}\,L&\text{ if }\|f\|_1\ge\frac{\sqrt d}{d+1}\,L. \end{aligned} \right. \end{align*}
Remark 1: Obviously, the inequality in (1) will turn into the equality if we choose $f=h_a$ with $a=0$, and then the inequality in (3) will turn into the equality as well. Moreover, the inequality in (2) will change the direction if we replace $v/\sqrt d$ in (2) by $v$. Therefore, the bound $B_0(\|f\|_1,L)$ is optimal up to a factor depending only on $d$. It also follows that the bound $B_0(\|f\|_1,L)$ is exact when $d=1$, in which case $B_0(\|f\|_1,L)$ is
exactly the same as the exact upper bound on $\|f\|_\infty$ presented in the other answer of mine on this web page (previously obtained somewhat differently).
Remark 2: We have $\|f\|_1\le\|f\|_2$, since the Lebesgue measure on $\Om$ is a probability measure. Also, $B_0(\cdot,L)$ is nondecreasing. So, $$\|f\|_\infty\le B_0(\|f\|_1,L)\le B_0(\|f\|_2,L).$$
Remark 3: One can show that $c_d\le\sqrt{2d}$ for all natural $d$.
Remark 4: It follows from Remark 3 that \begin{align*} \|f\|_\infty&\le B_1(\|f\|_1,L) \\ &:=\left\{\begin{aligned} \sqrt{2d}\, L^{d/(d+1)}\|f\|_1^{1/(d+1)}&\text{ if }\|f\|_1<\frac{\sqrt d}{d+1}\,L, \\ %\|f\|_1&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\ (d+1)\|f\|_1&\text{ if }\|f\|_1\ge\frac{\sqrt d}{d+1}\,L. \end{aligned} \right. \end{align*} Note that $B_1(\|f\|_1,L)$ differs from $B_0(\|f\|_1,L)$ by, at most, a factor depending only on $d$. So, in view of Remark 1, the bound $B_1(\|f\|_1,L)$ is optimal as well up to a factor depending only on $d$. Note also that the exponent of $\|f\|_1$ in the bound $B_1(\|f\|_1,L)$ is $1/(d+1)$ if $\|f\|_1$ is not too large as compared with $L$, and this exponent is $1$ otherwise.
In view of Remark 2, we also have $$\|f\|_\infty\le B_1(\|f\|_2,L).$$
Remark 5: As shown in Willie Wong's comment, if we had a bound on $\|f\|_\infty$ of the form $C(d)L^a\|f\|_1^b$, then the only possible values for $a$ and $b$ would be $d/(d+1)$ and $1/(d+1)$, respectively. However, in view of Remark 4, it is clear that such a bound on $\|f\|_\infty$ is impossible: the exponent of $\|f\|_1$ cannot be greater than $1/(d+1)$ for values of $\|f\|_1$ not too large as compared with $L$, and this exponent cannot be less than $1$ for values of $\|f\|_1$ large enough as compared with $L$.