On summability of functions in weighted Lebesgue spaces

Question

I consider a (continuous) function $V \colon \mathbb{R}^n \to \mathbb{R}$ such that $V \geq 1$ and $\lim_{|x|\to +\infty} V(x)=+\infty$. Let $u$ be a measurable function on $\mathbb{R}^n$ such that $$\int V(x) |u(x)|^2 \, dx < \infty.$$

Let now $p>2$ be a real number. Under what (general) assumptions on $V$ and possibly on $u$ can we conclude that $$\int V(x) |u(x)|^p \, dx$$ is also finite?

It seems to me that some additional condition must be imposed, since $p>2$. I do not know how to use the condition that $V$ diverges at infinity (and maybe this condition is useless here). Any idea or suggestion?

The case $V \to 0$ at infinity seems to be more popular in the literature, but I am working in the opposite setting.

Answer 1

So thanks to @SmallDeviation we now know that the original proof does not work, you have to add supplementary conditions on the asymptotic behavior of $u$ since measurable functions are a very wide set of functions, there are critical errors. I have various ideas of condition but finding an explicit necessary and sufficient condition seems impossible.

Here is a suitable counterexample: $$V = \|x\| + 1$$ $$u = \sum_{n >0} n \cdot 1_{\left[n, n + \frac{1}{n^{4 +\epsilon}}\right]}$$ with $\epsilon >0$.

The end was pure garbage thus I suppressed it (I only leave my answer for the counter example, however $u \to 0$ is a cool condition to add

Answer 2

Since the question is pretty broad, i do not know whether my answer will be satisfying to you. Please let me know whether the following fits your criteria or whether you want a different approach:

My impression is that one would need to impose some "growth condition" on $V$ together with some "moment condition" on $u$, in the following sense:

Asssume there exists some $\gamma>0$ such that $$\lim_{|x|\to\infty} \frac{V(x)}{|x|^\gamma}=0,$$ which intuitively means that $V$ does not grow faster that $|x|^\gamma$. Now let's also assume (and this is an assumption for a fixed $p$) that $u$ satisfies the moment condition $$\int |x|^{2\gamma}\cdot |u(x)|^{2(p-1)}dx<\infty.$$ Then by a simple application of Cauchy-Schwartz together with the realization that $\int |u(x)|^2dx$ must be finite, due to the fact that $V\geq 1$, we get: $$\begin{align*}\int V(x)|u(x)|^pdx&\leq C\int |x|^\gamma|u(x)|^pdx = C\int |x|^\gamma |u(x)|^{p-1}\cdot|u(x)|dx \\ (\text{Cauchy-Schwartz})&\leq\bigg(\int |x|^{2\gamma}|u(x)|^{2(p-1)}dx\bigg)^{1/2}\bigg(\int |u(x)|^2 dx\bigg)^{1/2}<\infty.\end{align*}$$ Alternatively, one could derive similar conditions by using Hölder's inequality instead of the Cauchy-Schwartz inequality. It may be worthwile to try Hölder's inequality with different exponents to see if you can get a sufficient condition that suits your problem. I.e. you can exchange how strong either the "growth condition" on $V$ is or how strong the integrability condition on $u$ is depending on which exponents you choose.