Weak convergence, $\displaystyle f\in L^p(\Bbb R^d) $ and $g\in L^q(\Bbb R^d) $ then, $\lim_{j\to\infty} \int_{\Bbb R^d}f_j(x)g(x)dx = 0$

Question

Assume $f\in L^p(\Bbb R^d) $ and $g\in L^q(\Bbb R^d) $ Where, $1<p<\infty$ and $1<q<\infty$ are dual ecxponents namely, $$\frac1p+\frac1q =1$$ Then for every $s\in\Bbb R$ such that, $sp\le d$ show that, $$\lim_{j\to\infty} \int_{\Bbb R^d}f_j(\mathbf{x})g(\mathbf{x})\mathrm{d}\mathbf{x} = 0$$

Where, $$f_j(\mathbf{x}) = j^sf(j\mathbf{x})~~~$$

My attempt,

I Applied Holder inequality, and use the change of variables $\mathbf{u} =j\mathbf{x}$ to get the following,

\begin{align}\left|\int_{\Bbb R^d}f_j(\mathbf{x})g(\mathbf{x})\mathrm{d}\mathbf{x}\right| &\le \left(\int_{\Bbb R^d}|f_j(\mathbf{x})|^p\mathrm{d}\mathbf{x}\right)^{1/p} \left(\int_{\Bbb R^d}|g(\mathbf{x})|^q\mathrm{d}\mathbf{x}\right)^{1/q}\\&=\frac{1}{j^{\frac dp-s}}\left(\int_{\Bbb R^d}|f(\mathbf{x})|^p\mathrm{d}\mathbf{x}\right)^{1/p} \left(\int_{\Bbb R^d}|g(\mathbf{x})|^q\mathrm{d}\mathbf{x}\right)^{1/q} \\&=\frac{1}{j^{\frac dp-s}}\|f\|_{p}\|f\|_{q}\end{align}

If I suppose that, $\color{blue}{d>sp}$ then $\frac{1}{j^{\frac dp-s}}\to 0$ and the result follows.

Question: How do I prove the case where $sp= d$?

Answer 1

This is just a reminiscent of the this: How to prove that, $\lim\limits_{k \rightarrow \infty} \int_{\Bbb{R}} f_k(x) g(x) dm(x) = 0$.

If $sp =d$ then, Proceed as follows:

For, $\varepsilon>0$ we know that there exist $g_\varepsilon\in C_c(\Bbb R)$ a continuous function of compact support such that, $$ \|g-g_\varepsilon\|_q\le \varepsilon \|f\|_p^{-1} $$ and for this choice of $g_\varepsilon$ we can also find $f_\varepsilon\in C_c(\Bbb R)$ such that

$$ \|f-f_\varepsilon\|_p\le \varepsilon \|g_\varepsilon\|_q^{-1}$$

Proceeding like here with Cauchy Schwartz inequality replaced by Holder inequality, with $d=sp$, one easily arrives at

$$\left|\int_\mathbb R f_j(x) g(x) dx\right| \le\left[\varepsilon+\varepsilon j^{s-d}+ j^{s-d}\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{j}) dm(x)\right| \right]\\ = \left[\varepsilon+\varepsilon j^{s(1-p)}+ j^{s(1-p)}\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{j}) dm(x)\right| \right]$$

Moreover, $f_\varepsilon, g_\varepsilon$ are continuous functions with compact support, we get $$\color{blue}{\lim_{j\to \infty}\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{j}) dx=\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(0) dx}$$

Whence, $$ \limsup_{j\to \infty} \left|\int_\mathbb R f_j(x) g(x) dx\right| \\ \le \limsup_{j\to \infty} \left[\varepsilon+\varepsilon j^{s(1-p)}+ j^{s(1-p)}\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{j}) dm(x)\right| \right]=\varepsilon $$

Thus, for all $\varepsilon>0,$ we have, $$ \limsup_{j\to \infty} \left|\int_\mathbb R f_j(x) g(x) dx\right| \le \varepsilon $$ letting $\varepsilon\to 0$ yields the require result.

$$ \color{red}{\lim_{j\to \infty} \int_\mathbb R f_j(x) g(x) dx =0.} $$

Answer 2

Note it's not true for $p=1$. Assume $p>1$. Come to think of it, it's also false for $p=\infty$. Assume $p<\infty$.

Let $\epsilon>0$. Choose $A>0$ so that $$\left(\int_{|x|\le A}|g(x)|^q\right)^{1/q}<\epsilon.$$(This is where $q<\infty$ is needed.) An explicit calculation shows that $$\left(\int_{|x|>A}|f_j(x)|^p\right)^{1/p}\to0.$$(This uses $p<\infty$.)

Edit: What's above will look like a solution to many readers. Others may be wondering how those two lines prove what's asserted. The (standard) argument goes like so: Assume $\epsilon$ and $A$ are as above. Choose $N$ so that $$\left(\int_{|x|>A}|f_j(x)|^p\right)^{1/p}<\epsilon\quad(j>N).$$ Noting that $||f_j||_p=||f||_p$ (since $sp=d$), for $j>N$ we have $$\left|\int f_jg\right|\le\left|\int_{|x\le A}f_jg\right|+\left|\int_{|x|>A}f_jg\right|\le||f||_p\epsilon+\epsilon||g||_q.$$