Suppose I have $\|f\_n\|\_{2}^2=\int\_{S}f\_n(x)^2dx\rightarrow 0$ over a compact set $S\subset R^d$, and $\{f\_n\}$ is $1$-Lipschitz and $1$-smooth. What kind of extra condition can I add on $S$ so that $\|f\_n\|\_{\infty}\rightarrow 0$? 


If $S$ is $R^d$, no extra condition is needed, since for any $|f(x)|>0$ we can always construct a neighborhood $B(x,r)\subset S$ for some $r=\Theta(|f(x)|)$ and the squared integral over $B(x,r)$ is $\Theta(|f(x)|^{d+2})$. However, for a compact set it does not necessarily holds that $B(x,r)\subset S$, so I guess some regularization condition can be added to make the convergence still hold.



How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?
It suffices that every neighborhood of every point in $S$ have a nonzero Lebesgue measure, that is, 
$$g\_r(x):=|S\cap B\_r(x)|>0\tag{1}
$$
for all $x\in S$ and all real $r>0$, where $B\_r(x):=\{y\in\mathbb R^d\colon\|y-x\|<r\}$, $\|\cdot\|$ is the Euclidean norm, and $|\cdot|$ is the Lebesgue measure. 


Indeed, it easy to see that for each $r>0$ the function $g\_r$ is continuous on the compact set $S$ and hence 
$$h\_r:=\min\_{x\in S}g\_r(x)=\min\_{x\in S}|S\cap B\_r(x)|>0, 
$$
and this is the crucial point. 


Suppose now that $\|f\|\_\infty\ge c$ for some real $c>0$. Then $|f(x\_f)|\ge c$ for some $x\_f\in S$ and hence $|f|\ge c/2$ on $S\cap B\_{r\_c}(x\_f)$, where $r\_c=\frac c{2L}>0$ and $L$ is the Lipschitz constant (one and the same for all the functions $f$ you want to deal with). So, 
$$\|f\|\_2\ge\frac c2\,\sqrt{|S\cap B\_{r\_c}(x\_f)|}\ge\frac c2\,\sqrt{h\_{r\_c}}>0. \tag{2}
$$
Thus, if $\|f\|\_2\to0$, then $\|f\|\_\infty\to0$. 


Indeed, if (say) for a sequence $(f\_n)$ we have $\|f\_n\|\_\infty\not\to0$, then, passing to a subsequence, without loss of generality we may assume that $\|f\_n\|\_\infty\ge c$ for some $c>0$ and all $n$. But then, by (2), 
$\|f\_n\|\_2\ge b>0$ for all $n$, where $b:=\frac c2\,\sqrt{h\_{r\_c}}>0$ -- so that $\|f\_n\|\_2\not\to0$. 




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It is also easy to see that condition (1) is, not only sufficient, but also necessary. Indeed, suppose that (1) fails to hold for some $x\in S$ and some real $r>0$. For all $y\in S$, let $f(y)$ be defined as the shortest distance from $y$ to $\mathbb R^d\setminus B\_r(x)$. Then $f$ is $1$-Lipschitz and $\|f\|\_2=0$, whereas $\|f\|\_\infty=r>0$. 


