$\lVert f_n \rVert_1$ convergent and $f_n \to 0$ 'everywhere' implies $\lVert f_n \rVert_1 \to 0$??

Question

Let $\{f_n \}$ be a sequence of smooth integrable functions on $\mathbb{R}$ such that the $L^1$ norm $\lVert f_n \rVert_1$ is convergent to some finite value as $n \to \infty$.

If we assume that $f_n \to 0$ pointwise on $\mathbb{R}$ (note that it is NOT a.s convergence, so I think I can rule out the case of point mass), then is it true that $\lVert f_n \rVert_1 \to 0$?

It seems trickier than expected..Could anyone please hep me?

edit) Ok, thank you for the answer. What if I change the condition 'pointwise' to 'uniformly'? Does this make any difference on $\mathbb{R}$?

Answer 1

No. Here is an example where the $f_n$ are not smooth. You can modify it as you see fit.

Let $f_n(x) = \chi_{[n,n+1]}(x)$. Then $\|f_n\|_1 = 1$ for all $n$, yet $f_n(x) \to 0$ everywhere.

Answer 2

If you ask a mathematics student using the language of the Lebesgue integral, most students will consider the problem to be sophisticated and look for some baroque kind of answer.

But this problem is an elementary calculus problem. Pointwise convergence does not imply convergence of integrals. Didn't we learn that? Wasn't that the reason why we had to learn uniform convergence long ago?

Express it, however, as pointwise convergence does not imply convergence in $L^1$ and it seems we are not in Kansas any longer Toto.

But you are.

Take a look at this answer: Prove that $\lim_{n \rightarrow \infty} \int_0^1 f_n(x)dx \ne \int_0^1\lim_{n \rightarrow \infty}f_n(x) dx$

Too lazy for a click of the mouse? Here it is in the form of a calculus question which immediately answers your question and does it in a compact interval:

Problem. If $f_n(x)=nxe^{-nx^2}~\forall~n=1,2,\cdots$ show that $f_n(x)\to 0$ for all $0\leq x \leq 1$ but that $\lim_{n\to\infty }\int_0^1 f_n(x)dx=\frac{1}{2}.$

One of the answers even does all the calculus steps for you, steps which advanced students hoped were far behind them:

$$\int_0^1 f_n(x)dx=n\int_0^1 xe^{-nx^2}dx=n\left[-\frac{1}{2n}e^{-nx^2}\right]_0^1=n\left(\frac{1}{2n}-\frac{1}{2n}e^{-n}\right)=\frac{1}{2}-\frac{1}{2}e^{-n}\to \frac12.$$

You can integrate on $[0,\infty)$ too if you wish.

P.S. Your added question about uniform convergence? On a compact interval a uniformly convergent sequence of bounded functions would be uniformly bounded. So Lebesgue's bounded convergence theorem will give you an answer. On an unbounded interval, you would want to use the Lebesgue dominated convergence theorem and uniform convergence won't assist in that.