# $f \in W^{m,2} ( \Bbb R^n) \cap C^{\infty} (\Bbb R^n )$ then $\| f \|_{L^\infty} < \infty$?

If $f \in W^{m,2} ( \Bbb R^n) \cap C^{\infty} (\Bbb R^n )$ for $m = 0,1,\cdots$, then can I conclude that $$\| f \|_{L^\infty} < \infty \;?$$

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Not for $m=0$ and $n=1$ for instance. There are smooth functions in $L^2$ which are not bounded. – Sam Oct 11 '12 at 19:55
You should clarify: is $f\in W^{m,2}$ true for any $m\geq 0$ or some $m\geq 0$? – Davide Giraudo Oct 11 '12 at 22:11

We have $$\frac 12\left|f(x)^2-f(0)^2\right|=\left|\int_0^xf'(t)f(t)dt\right|\leq \lVert f\rVert_{L^2}\lVert f'\rVert_{L^2},$$ hence $|f(x)^2-f(0)^2|\leq 2\lVert f\rVert_{L^2}\lVert f'\rVert_{L^2}$ and $$|f(x)|^2\leq |f(0)|^2+2\lVert f\rVert_{L^2}\lVert f'\rVert_{L^2},$$ so it works in dimension $1$ if $f\in W^{1,2}$. It also works on other dimensions, as $$|f(x_1,\dots,x_n)^2-f(0)^2|=2\int_{0}^{x_n}\partial_nf(x_1,\ldots,x_{n-1},t)\cdot f(x_1,\ldots,x_{n-1},t)dt+\dots,$$ writing $|f(x)^2-f(0)^2|=\sum_{j=1^n}|f(x^{(j)})-f(x^{(j-1)}|$, where $x^{(j)}=(x_1,\dots,x_j,0,\dots,0)$ and $x^{(0)}=(0,\dots,0)$.