# Thread: Verify a Proof about L2 Norm

1. ## Verify a Proof about L2 Norm

I'm working through some suggested problems and one asks to show that if $f_{n} \rightarrow f$ uniformly on a compact interval $I$, and each $f_{n}$ continuous on $I$, then $\displaystyle \lim_{n \rightarrow \infty} || f_{n} - f|| = 0$, where ||...|| is the $L^{2}$ norm. I first want to verify that the following is sound: If $\displaystyle \lim_{n \rightarrow \infty} \int_{I} (f_{n} - f)^{2} = 0$ then $\displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$. 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 $n$, for all $x \in I$ $|f_{n}(x) - f(x)| < \varepsilon$, and since $I$ is compact it's therefore bounded, let's say the length of the interval is bounded by $M$. So $\displaystyle \int_{I} |f_{n} - f|^{2} < M\varepsilon$. But at no point did I seem to use continuity of $f_{n}$...

2. Originally Posted by ragnar
I'm working through some suggested problems and one asks to show that if $f_{n} \rightarrow f$ uniformly on a compact interval $I$, and each $f_{n}$ continuous on $I$, then $\displaystyle \lim_{n \rightarrow \infty} || f_{n} - f|| = 0$, where ||...|| is the $L^{2}$ norm. I first want to verify that the following is sound: If $\displaystyle \lim_{n \rightarrow \infty} \int_{I} (f_{n} - f)^{2} = 0$ then $\displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$. 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 $n$, for all $x \in I$ $|f_{n}(x) - f(x)| < \varepsilon$, and since $I$ is compact it's therefore bounded, let's say the length of the interval is bounded by $M$. So $\displaystyle \int_{I} |f_{n} - f|^{2} < M\varepsilon$. But at no point did I seem to use continuity of $f_{n}$...
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 $\lambda (I) < \infty$ ( $I$ need not be an interval) and that $f_n \rightarrow f$ uniformly on $I$ and your argument still works.

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