$f_n\to f$ in $L^2$ implies $(f_n)^k\to f^k$ in the sense of distributions for all $k\in\mathbb{N}$?

Question

I have a very naive question: Let $\{f_n\}_{n\in\mathbb{N}}\subset L^2(\mathbb{R})$ a sequence of functions satisfying $f_n\to f$ in $L^2$ for some $f\in L^2(\mathbb{R})$. Does this implies that, for all $k\in\mathbb{N}$, we have $$ (f_n)^k\to f^k \quad \hbox{in the sense of distributions }? $$ Is this somehow direct? or maybe directly false? I have the feeling that it might be false since, for example, even on compact sets we could have functions belonging to $L^2$ such that they are not in $L^3$. However, with this naive intuition we could guess that the above convergence should hold for $k=1$, right?

Answer 1

This is false for $k>2$ as you already mentioned. For $k\in [1,2]$ it is true. Let $\phi$ be smooth with compact support, then by Hoelder inequality $$ \int_{\mathbb R} (f_n-f)^k \phi = \int_{supp \ \phi} (f_n-f)^k \phi \le \|\phi\|_{L^\infty} \|f_n-f\|_{L^2}^k |supp \ \phi|^{1-\frac k2} = c(\phi) \|f_n-f\|_{L^2}^k $$ with $\frac1q +\frac1k = \frac12$. Here, it is important that the support of $\phi$ is compact and so has finite measure.

Answer 2

Hint

Let $\varphi $ is smooth. By Cauchy-Schwarz, $$\left|\int_{\mathbb R}(f_n-f)\varphi ^{(k)}\right|\leq \|f_n-f\|_{L^2}\|\varphi ^{(k)}\|_{L^2}.$$