Some of the books that discuss convergence say that uniform convergence implies $L^2$ convergence and $L^2$ convergence implies $L^1$ convergence, both while taken over a bounded interval I. While I understand how that could be true intuitively, I'm struggling to see the proof of that. Any ideas?
EDIT: For Uniform Convergence implies $L^2$ convergence, uniform convergence and interchange of limits means $\lim\limits_{n\rightarrow\infty} \int_{a}^{b}(f(x)-f_n(x)) = 0$ and from there, I'm not sure what to do?
For $L^2$ convergence, I don't actually know where to get started either.
Best Answer
If $f_n$ converges to $f$ uniformly (i.e. $\sup_{x \in I}|f_n(x)-f(x)| \to 0$), then $$\|f_n-f\|^2_2=\int_I |f_n(x)-f(x)|^2\,dx \leq m(I) \left(\sup_{x \in I}|f_n(x)-f(x)|\right)^2 \to 0,$$ so $f_n$ converges to $f$ in $L^2$.
If $f_n$ converges to $f$ in $L^2(I)$, then by Hölder inequality we have $$ \|f_n-f\|_1=\int_I |f_n(x)-f(x)|\,dx \leq (m(I))^{\frac12} \|f_n-f\|_2 \to 0, $$
so that $f_n$ goes to $f$ in $L^1$.
Note that these results (with the same proof) hold in a much more general context, as pointed out in the other answer, I tried to give the most "hands on" proof possible.