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Given $u \in L^2(\mathbb{R}^N)$ and $\nabla u \in L^\infty(\mathbb{R}^N)$, is $u \in L^\infty(\mathbb{R}^N)$?

Can I use Morrey's inequality? $$|u(x) - u(y)| \leq C_N \|\nabla u\|_\infty |x-y| \text{ a.e. }$$

Although this is an inequality for $u\in W^{1,\infty}(\mathbb{R}^N)$, but I believe in the proof of this inequality, we do not use the fact that $u\in L^\infty(\mathbb{R}^N)$, normally it is done with a density with $u\in C_c^\infty(\mathbb{R}^N)$.

If we have this inequality, then $u$ is Lipschitz continuous, it does not allow $u$ to have any spikes on $\mathbb{R}^N$ and $u$ can not grow to unbounded outside of any large ball because $u\in L^2(\mathbb{R}^N)$.

Thank you very much!

Xiao
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Yes, your idea is correct. Every function $u$ with weak gradient $\nabla u\in L^\infty(\mathbb{R}^n)$ has a Lipschitz representative (e.g., by the argument in the second paragraph here). Being Lipschitz and in $L^2$, it has to be essentially bounded. Indeed, if $|u(x_0)|>M$, then $|u|>M/2$ on the ball of radius $M/(2L)$ centered at $x_0$, where $L$ is the Lipschitz constant of $u$. Hence, $$ \frac{M^2}{4L^2}\le \int_{\mathbb{R}^n}u^2 $$ which gives a bound on $M$.

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