Continuous function sequence that converges in $L^1$ but not in $L^2$

Question

Hello I am trying either to find an example that this is false or a proof that this is true.

Let $f_n$ be a sequence of continuous functions then if $f_n$ converges to f in $L^1$ then $f_n$ converges to f in $L^2$.

I am pretty sure this is false, since its false for a sequence of non-continuous functions (i.e. $f_n(x)= 1/\sqrt{x}$ if $x\in (0,1/n)$ and values $0$ elsewhere.)

Cannot think of an counterexample though.

Answer 1

Try something like $f_n = a_n1_{[0,1/n]}$ and consider convergence to zero. Then

• $\|f_n\|_{L^1} = a_n n^{-1}$,
• $\|f_n\|_{L^2} = a_n n^{-1/2}$,

so the $L^2$ norm is larger than the $L^1$ norm for these functions. If the coefficient sequence $a_n = n^{1/2}$, then $\|f_n\|_{L^2} = 1$ for all $n$, but $\|f_n\|_{L^1} = n^{-1/2} \to 0$.

This sequence can be modified to be continuous by considering an appropriate tent or "triangle" function in place of the discontinuous step function with the same conclusion.

Answer 2

There are many counterexamples to this but some of the easier ones are (built from) polynomials. For this we distinguish two cases:

1. Functions on finite intervals, i.e. $(a,b)$ where $-\infty<a<b<\infty$.

Here one can choose something as simple as $f_n(x):=\sqrt n(\frac{x-a}{b-a})^n$. These functions are not only continuous but even analytic, and their sequence converges to the zero function in $L^1([a,b])$ $$ \|f_n-0\|_1=\int_a^b \sqrt n\Big(\frac{x-a}{b-a}\Big)^n=\frac{\sqrt n}{n+1}\Big[\Big(\frac{x-a}{b-a}\Big)^{n+1}\Big]_a^b=\frac{\sqrt{n}}{n+1}\to 0\text{ as }n\to\infty $$ but not in $L^2([a,b])$ $$ \|f_n-0\|_2=\int_a^b n\Big(\frac{x-a}{b-a}\Big)^{2n}=\frac{n}{2n+1}\Big[\Big(\frac{x-a}{b-a}\Big)^{2n+1}\Big]_a^b =\frac{n}{2n+1}\not\to 0\text{ as }n\to\infty\,. $$

2. Functions on infinite intervals, i.e. $(a,b)$, $a<b$ where $a=-\infty$ or $b=\infty$ (or both)

Because the functions from the previous example satisfy $f_n(a)=0$ they can be continuously extended to $(-\infty,b)$ by $f_n(x):=0$ for all $x\leq a$; this "takes us to infinity" and on the other side of the function we could for example just mirror it. Assume w.l.o.g. that $(-1,1)\subset(a,b)$ (else we can just shift the following function) and define \begin{align*} f_n:(a,b)&\to\mathbb R\\ x&\mapsto \begin{cases} \sqrt n(1-|x|)^n&x\in[-1,1]\\0&\text{else} \end{cases} \end{align*} for all $n\to\infty$. Again each $f_n$ is continuous and $\|f_n-f\|_1\to 0$ but $\|f_n-f\|_2\to 1\neq 0$ as $n\to\infty$.