The Riesz-Fischer Theorem implies that Lp-convergence implies pointwise a.e. convergence of a subsequence.
There is an example that shows that the converse may not be true...
Let E = [0, 1], $1 \leq p < \infty$ and $f_n = n^{1/p}\chi_{(0,1/n]}$. Then $f_n \rightarrow 0$ pointwise but $||fn − 0||_p = 1$ and so $f_n$ does not converge to 0 with respect to the Lp norm.
I'm a bit confused by how $||fn − 0||_p = 1$ is true. So if I try to compute the norm...
$||fn − 0||_p = [\int(f_n-0)^p]^{1/p} = [\int n\chi_{(0,1/n]}]^{1/p}$.
We know that 1/n goes to zero as n approaches zero. So eventually we will have $[\int n\chi_{(0,0]}]^{1/p}$, and that's just zero, right?
I know I must be doing something wrong, but I couldn't really figure out what.
Thanks in advance
You cannot take the limit $n\to \infty$ for only one term. It's like if you wrote that $$n=n^2\frac 1n\underset{n\to\infty}{\rightarrow} n^2\cdot 0=0.$$ The measure of $[0,n^{-1})$ is $n^{—1}$, hence $\lVert f_n\rVert_p=1$ for each $n$.