# Verify a Proof about L2 Norm

• Aug 3rd 2010, 03:25 PM
ragnar
Verify a Proof about L2 Norm
I'm working through some suggested problems and one asks to show that if $\displaystyle f_{n} \rightarrow f$ uniformly on a compact interval $\displaystyle I$, and each $\displaystyle f_{n}$ continuous on $\displaystyle I$, then $\displaystyle \displaystyle \lim_{n \rightarrow \infty} || f_{n} - f|| = 0$, where ||...|| is the $\displaystyle L^{2}$ norm. I first want to verify that the following is sound: If $\displaystyle \displaystyle \lim_{n \rightarrow \infty} \int_{I} (f_{n} - f)^{2} = 0$ then $\displaystyle \displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$. My main worry is moving the limit outside of the square root in this argument.

Anyway, given that this is sufficient, then we know that for sufficiently large $\displaystyle n$, for all $\displaystyle x \in I$ $\displaystyle |f_{n}(x) - f(x)| < \varepsilon$, and since $\displaystyle I$ is compact it's therefore bounded, let's say the length of the interval is bounded by $\displaystyle M$. So $\displaystyle \displaystyle \int_{I} |f_{n} - f|^{2} < M\varepsilon$. But at no point did I seem to use continuity of $\displaystyle f_{n}$...
• Aug 3rd 2010, 03:44 PM
Jose27
Quote:

Originally Posted by ragnar
I'm working through some suggested problems and one asks to show that if $\displaystyle f_{n} \rightarrow f$ uniformly on a compact interval $\displaystyle I$, and each $\displaystyle f_{n}$ continuous on $\displaystyle I$, then $\displaystyle \displaystyle \lim_{n \rightarrow \infty} || f_{n} - f|| = 0$, where ||...|| is the $\displaystyle L^{2}$ norm. I first want to verify that the following is sound: If $\displaystyle \displaystyle \lim_{n \rightarrow \infty} \int_{I} (f_{n} - f)^{2} = 0$ then $\displaystyle \displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$. My main worry is moving the limit outside of the square root in this argument.

Anyway, given that this is sufficient, then we know that for sufficiently large $\displaystyle n$, for all $\displaystyle x \in I$ $\displaystyle |f_{n}(x) - f(x)| < \varepsilon$, and since $\displaystyle I$ is compact it's therefore bounded, let's say the length of the interval is bounded by $\displaystyle M$. So $\displaystyle \displaystyle \int_{I} |f_{n} - f|^{2} < M\varepsilon$. But at no point did I seem to use continuity of $\displaystyle f_{n}$...

Your argumentts look fine to me. Notice we don't need I to be compact or the f_n to be continous. Assume only that $\displaystyle \lambda (I) < \infty$ ($\displaystyle I$ need not be an interval) and that $\displaystyle f_n \rightarrow f$ uniformly on $\displaystyle I$ and your argument still works.

Edit: Another argument that works is using Dominated convergence in $\displaystyle L^2$ to the functions $\displaystyle |f_n|\leq |f|+\varepsilon$ for $\displaystyle n\geq N(\varepsilon )$