1
$\begingroup$

Suppose I have a sequence $\{f_{n}\}_{n\in \mathbb{N}} \subset H^{1}(\mathbb{R}^{d})$ which converges weakly to $f$ in $H^{1}(\mathbb{R}^{d})$, in the sense that $\langle f_{n},\varphi \rangle_{L^{2}}+\langle \nabla f_{n},\nabla \varphi\rangle \to \langle f, \varphi\rangle_{L^{2}}+\langle \nabla f, \nabla \varphi\rangle_{L^{2}}$ for every function $\varphi$. I know that this implies, in particular, that $f_{n} \rightharpoonup f$ in $L^{2}(\mathbb{R}^{d})$, in the sense that $\langle f_{n},h\rangle_{L^{2}} \to \langle f, h\rangle_{L^{2}}$ for every $h \in L^{2}(\mathbb{R}^{d})$.

Suppose $g \in L^{p}(\mathbb{R}^{d})$ is fixed for $p > \max\{\frac{d}{2},1\}$. Can something be said about the following convergence? $$\lim_{n\to \infty}\int_{\mathbb{R}^{d}}|f_{n}(x)|^{2}g(x)dx \to \int_{\mathbb{R}^{d}}|f(x)|^{2}g(x)dx$$ I believe the answer is yes, the convergence holds. However, not sure how to prove it. My guess would be something like in the lines of using: \begin{align}\int_{\mathbb{R}^{d}}|g(x)|(|f_{n}(x)|^{2}-|f(x)|^{2})dx &= \int_{\mathbb{R}^{d}}|g(x)|(|f_{n}(x)|-|f(x)|)(|f_{n}(x)|+|f(x)|)dx \\ &\le \int_{\mathbb{R}^{d}}|g(x)||f_{n}(x)-f(x)|(|f_{n}(x)|+|f(x)|)dx \end{align} By Banach-Steinhaus, we can bound $|f_{n}(x)|$ by a constant and, taking the constant large enough, we can even bound $|f_{n}(x)|+|f(x)| \le C(1+|f(x)|)$. So maybe it is just a matter of dealing with the $g$ function somehow?

$\endgroup$

1 Answer 1

4
$\begingroup$

This is true. Assume for example that $d \geq 3$. Since $(f_n)$ is bounded in $H^1$, it is bounded in $L^q$ for $2 \leq q \leq \frac{2d}{d-2}$, by Sobolev embedding. Moreover $f_n \to f$ strongly in $L^q(B_R)$, $2 \leq q <\frac{2d}{d-2}$, for every $R>0$, by the compactness of Sobolev embedding when restricted to balls and then $|f_n|^2 \to |f|^2$ in $L^q(B_R)$ if $1 \leq q <\frac{d}{d-2}$. This implies that $$\int_{\mathbb R^d}|f_n|^2g \to \int_{\mathbb R^d}|f|^2g $$ for every continuous $g$ with compact support and then, by density and boundedness of $\||f_n|^2\|_q$, for every $g \in L^{q'}$.

$\endgroup$
2
  • $\begingroup$ Thanks! You say $(f_{n})$ is bounded in $H^{1}$ because it is weakly convergent? Is it easy to see? $\endgroup$ Commented Sep 20, 2023 at 8:10
  • $\begingroup$ Yes, it is a standard fact which holds in any Banach space. If a sequence converges weakly, then it is weakly bounded and then strongly bounded, by Banach-Steinhaus. $\endgroup$ Commented Sep 20, 2023 at 8:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .