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$\DeclareMathOperator\Lip{Lip}$This question arose when I read Godefroy and Lerner - Some natural subspaces and quotient spaces of $L^1$.

Let $\Lip_0(\mathbb R^n)$ be the space of Lipschitz functions $f:\mathbb R^d\to\mathbb R$ vanishing at the origin, $f(0)=0$. It is known from the above paper that, endowed with the norm $\|f\|_{\Lip}\mathrel{:=}\|\nabla f\|_{\infty}$, $\big(\Lip_0(\mathbb R^n), \|\cdot\|_{\Lip}\big)$ is a Banach space. My question is, if $f^n$ converges to $f$ under the above norm, could we deduce $$\lim_{n\to\infty} \int_{\mathbb R^d}\big(f^n(x)-f(x)\big)u(x)dx = 0 ,$$ where $u:\mathbb R^d \to\mathbb R_+$ is a measurable function s.t. $$ \int_{\mathbb R^d}(1+\lvert x\rvert)u(x)dx <\infty.$$

This seems a trivial question, but I can not prove it rigorously for general dimensions. Is there any classical reference?

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Denote $g_n:=f-f_n$, so that $\lim_{n\rightarrow\infty}\|\nabla g_n\|_{\infty}=0$ . Note that by the fundamental theorem of calculus we have $$ g_n(x)=g_n(x)-g_n(0)=\int_0^1 \frac{d}{dt}(g_n(tx))dt=\int_0^1 x\cdot \nabla g_n(tx)\, dt, \qquad x\in\mathbb{R}^d. $$ Using the above formula we get \begin{align*} \lim_{n\to\infty} \left|\int_{\mathbb R^d} g_n(x)u(x)dx\right| &= \lim_{n\to\infty} \left|\int_{\mathbb R^d} \left(\int_0^1 x\cdot \nabla g_n(tx)\, dt\right)\, u(x)\,dx\right|\\ &\le \lim_{n\to\infty} \int_{\mathbb R^d} \left(\int_0^1 |x||\nabla g_n(tx)|\, dt\right)\, u(x)\,dx\\ &\le \lim_{n\to\infty} \|\nabla g_n\|_{\infty}\int_{\mathbb R^d} |x|\, u(x)\,dx=0. \end{align*}

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  • $\begingroup$ Many thanks. I forgot this argument :) $\endgroup$ – Neymar May 10 '20 at 13:33
  • $\begingroup$ You're welcome :) $\endgroup$ – Tony419 May 10 '20 at 14:27
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Here is an alternative answer (also based on the control of the growth at infinity): simply use Lebesgue's Dominated Convergence theorem: Note first that the convergence $\|f_n-f\|\to 0$ in your $Lip_0(\mathbb R^d)$ space immediately implies pointwise a.e. convergence, $$ f_n(x)u(x)\to f(x)u(x) \qquad a.e. $$ In order to apply the DCT we only need a dominating $L^1$ bound. For this note that the Lipschitz norm controls the growth at infinity, hence $ |f_n(x)|\leq \|f_n\|\,|x|\leq 2 \|f\|\, |x| \qquad \forall x$ uniformaly in $n$. In particular given your assumptions on $u$ we get $$ |f_n(x)\, u(x)|\leq 2 \|f\|\, |x|\, u(x)\in L^1 $$ and the result follows.

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  • $\begingroup$ Many thanks for the answer. $\endgroup$ – Neymar May 10 '20 at 13:33

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