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

If $S_n$ 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 integral over $B(x,r)$ is $\Theta(|f(x)|^{d+1})$. However, for a compact set it does not necessarily holds that $B(x,r)\subset S_n$, so I guess some regularization condition can be added to make the convergence still hold.

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.

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

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 integral over $B(x,r)$ is $\Theta(|f(x)|^{d+1})$. 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.

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

If $S_n$ 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 integral over $B(x,r)$ is $\Theta(|f(x)|^{d+1})$. However, for a compact set it does not necessarily holds that $B(x,r)\subset S_n$, so I guess some regularization condition can be added to make the convergence still hold.

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

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 integral over $B(x,r)$ is $\Theta(|f(x)|^{d+1})$. 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.

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

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 integral over $B(x,r)$ is $\Theta(|f(x)|^{d+1})$. 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.