# Uniform Convergence Implies $L^2$ Convergence and $L^2$ Convergence Implies $L^1$ Convergence

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.

• Have you tried anything? Nov 14, 2015 at 21:08
• Please point out which part of proof struggled you. Nov 14, 2015 at 21:11
• You should say you are on an interval $[a,b].$
– zhw.
Nov 15, 2015 at 0:13

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.

On a finite measure space $L^q$ convergence implies $L^p$ convergence for any $1\leq p \leq q \leq \infty$. Take $\infty>q>p$, so $q/p > 1$.This follows from Holder's inequality, $$||f||_p^p = \int |f|^p\,d\mu = \int |f|^p \cdot 1\,d\mu \leq \left(\int (|f|^{p})^{q/p}\,d\mu\right)^{p/q}\left(\int 1\,d\mu \right)^{1-p/q}= ||f||_q^p \mu(X)^{1-p/q}$$ so we see that if $f \to 0$ in $L^q$ then $f \to 0$ in $L^p$.

The case $q=\infty$ is easier, just using $|f| \leq ||f||_\infty$ almost everywhere, $$||f||_p^p = \int |f|^p\,d\mu \leq \mu(X) ||f||_\infty^p$$ and again the result follows.

• @AnirbanMukherjee I've double checked and I think it is correct as stated. $s:=q/p$ and $1/s + 1/r = 1$ implies that the exponent $1/r = 1-1/s = 1-p/q$. Jun 11, 2021 at 14:45
• You are correct. Thanks for checking. Jun 11, 2021 at 23:49