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.



Bounding supremum norm of Lipschitz function by L1 norm
$\newcommand\Om\Omega$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 $L$.


