Convergence in $\mathcal{S}'$ and $L^p$ spaces I expect that the following statement is true:

Let $p \in (0, \infty], f \in L^p(\mathbb{R}^n), \{f_j\} \subset L^1(\mathbb{R}^n)$ and $f_j \to f \in \mathcal{S}'(\mathbb{R}^n)$. In addition, if $\{f_j\}$ is $L^1$-bounded, then, $f \in L^1(\mathbb{R}^n)$.

I can neither prove nor disprove this claim.
I would be grateful if anyone could give me some advice.
My English is not very good. Please forgive me if there are any grammatical errors or confusing parts.
 A: Lets assume $p≥1$. $f_j\to f$ in $\mathcal S'$ means that $\int_{\Bbb R} f_j s\to\int_{\Bbb R} f s$ for all $s\in\mathcal S$, this condition can be weakened a bit, fix some $n$ and then you have:
$$\int_{[-n,n]}f_j(x) s(x)\,dx\to\int_{[-n,n]}f(x)s(x)\,dx$$
for all $s\in \mathcal S$. In particular the left-hand side is bounded above by $C\cdot \|s\|_\infty$ for $C:=\sup_{j}\|f_j\|_1$, so the right-hand side also is bounded above by this, ie:
$$\int_{[-n,n]}f(x) s(x)\,dx≤ C\,\|s\|_\infty$$
Three things:

*

*$L^1 ([-n,n])\supseteq L^p([-n,n])$ for $p\in[1,\infty)$, so $f\lvert_{[-n,n]}\in L^1([-n,n])$.

*The dual of $C([-n,n])$ includes $L^1([-n,n])$, so you can get the norm $\|f\lvert_{[-n,n]}\|_1$ by taking the supremum of $f[s]$ over all unit vectors $s$ in $C([-n,n])$.

*The restriction of $\mathcal S$ to $[-n,n]$ is dense in $C([-n,n])$ if you give it the norm $\|\cdot\|_\infty$. So in the above consideration you can just take the norm over the norm $≤1$ vectors in $\mathcal S$ rather than $C([-n,n])$.

Written out:
$$\|f\lvert_{[-n,n]}\|_1 = \sup_{s\in C[-n,n], \|s\|_\infty ≤1} f[s] =\sup_{s\in \mathcal S, \|s\|_\infty ≤1}\ \int_{[-n,n]}f(x)s(x)\,dx≤ C$$
And then:
$$\lim_n \int_{[-n,n]}|f(x)|\,dx ≤ C$$
giving that $f\in L^1$.
A: This is not true if you only assume $L^1$ boundedness of the sequence $\{ f_j\}$: Take any function $g\in L^1(\mathbb{R}^n)$ and set $f_j(x)=j^ng(jx)$. You can check that $f_j\to c\delta$ in $\mathcal{S}'$, where $c=\int_{\mathbb{R}^n} g$ and $\delta$ is the Dirac mass at the origin.
On the other hand, if you ask for $L^p$ boundedness for some $p>1$ the result is true by reflexivity of these spaces.
