Let $f \in W^{1,1}(\mathbb R^n)$. Suppose for each $k \in \mathbb Z_+$, $f_k \in W^{1, k} (\mathbb R^n)$, $\|f_k\|_{W^{1, k}(\mathbb R^n)} \leq 1$ and further $f_k \to f$ uniformly.
Question: Is it true that $f \in W^{1, \infty} (\mathbb R^n)$?
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- 2$\begingroup$ This is basically the argument that p-harmonic functions with the same boundary data converge to a Lipschitz minimizer. This isn't particularly useful in the scalar case because we have viscosity solutions, but it's very useful when viscosity solutions are unavailable, see Section 5 of arxiv.org/abs/2205.08250 $\endgroup$Aidan Backus– Aidan Backus2025-01-03 19:10:28 +00:00Commented Jan 3 at 19:10
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2 Answers
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3 For any test function $\varphi$ and any $k\in\mathbf{Z}_{>0}$ you have $$\left| \int f_k \varphi \right| \leq \|\varphi\|_{k'},$$ where $k'$ is the conjugate of $k$. Since $k'\rightarrow 1$ as $k\rightarrow +\infty$, the uniform convergence of $(f_k)_k$ towards $f$ implies $$\left|\int f \varphi\right| \leq \|\varphi\|_1,$$ which is a dual way to express that $f$ is essentially bounded by $1$. Similarly, $$ \left|\int f_k \nabla \varphi \right| = \left|\int\varphi \nabla f_k \right|\leq \|\varphi\|_{k'},$$ which just as above leads to $$ \left|\int f\nabla \varphi \right|\leq \|\varphi\|_{1}.$$ This last estimates tells you, again by duality, that the distributional gradient of $f$ is bounded so that $f$ is indeed lipschitz.
- $\begingroup$ At the end, I think one needs an argument why the distributional gradient of $f$ is even a function before one can conclude that this function is bounded. $\endgroup$Jochen Glueck– Jochen Glueck2025-01-03 08:43:50 +00:00Commented Jan 3 at 8:43
- 1$\begingroup$ @JochenGlueck, $f$ is assumed $W^{1,1}(\mathbf{R}^n)$. $\endgroup$Ayman Moussa– Ayman Moussa2025-01-03 09:48:32 +00:00Commented Jan 3 at 9:48
- 1$\begingroup$ But even, without it : any distribution $T\in\mathscr{D}'(\mathbf{R}^n)$ satisfying an estimate $|\langle T,\varphi\rangle|\lesssim \|\varphi\|_1$ for all test functions $\varphi$, is represented by a $L^\infty(\mathbf{R}^n)$ function (use the extension theorem for linear continuous maps defined on a dense subset and the Riesz identification of $L^1(\mathbf{R}^n)'$ as $L^\infty(\mathbf{R}^n)$). $\endgroup$Ayman Moussa– Ayman Moussa2025-01-03 09:51:41 +00:00Commented Jan 3 at 9:51
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2 By the assumption , for every ball $\Omega$ of Lebesgue measure $1/n$, and for every $1<p<\infty$, the sequence $(f_k)_k$ has $$\|f_k\|_{1,p, \Omega}\le \|f_k\|_{1,k, \mathbb R^n}\le 1$$ as soon as $k\ge p$. By weak compactness, and since it already converges uniformly to $f$, $f_k$ converges to $f$ weakly in $W^{1,p}(\Omega)$ and $\|f\|_{1,p, \Omega}\le 1$, for every $1<p<\infty$ and for every ball $\Omega $ of measure $1/n$. Then also $\|f\|_{1,\infty, \Omega}\le 1$ for every such ball, and then finally $\|f\|_{1,\infty, \mathbb R^n}\le 1.$
$$*$$ Details. Just recall that the $L^p$ norm $\|\cdot\|_{p,X}$ on a probability measure space $X$ is increasing w.r.to $p$, and converges to the $L^\infty$ norm as $p\to\infty$. In particular, for an open set $\Omega\subset\mathbb R^n$ of Lebesgue measure $1/n$, the norm of $W^{1,p}(\Omega)$, $\|f\|_{1,p,\Omega}:=\Big(\|f\|_{p,\Omega}^p+\sum_{i=1}^n\|\partial_i f\|_{p,\Omega}^p\Big)^{1/p}$ (seen as $L^p$ norm on a space $X:=\{0,\dots,n\}\times\Omega$ of measure $1$) is increasing w.r.to $p$ and converges to the $W^{1,\infty}$ norm on $\Omega$ $\|f\|_{1,\infty,\Omega}:=\max\big\{\|f\|_{\infty,\Omega}, \|\partial_1 f\|_{\infty,\Omega},\dots, \|\partial_n f\|_{\infty,\Omega} \big\}. $
- $\begingroup$ You mean converges weakly in $W^{1,p}$, I Guess. $\endgroup$Giorgio Metafune– Giorgio Metafune2025-01-03 19:29:41 +00:00Commented Jan 3 at 19:29
- $\begingroup$ yes thanks, “weakly” dropped, fixed :) $\endgroup$Pietro Majer– Pietro Majer2025-01-03 19:55:23 +00:00Commented Jan 3 at 19:55