Proof of L^2 Norm Property

**ragnar** · August 9th 2010, 06:07 PM

I am trying to prove that, if $\displaystyle \lim_{n \rightarrow \infty}||f_{n} - f|| = 0$ and $\lim_{n \rightarrow \infty} f_{n}(x) = g(x)$ almost everywhere on the I, for all functions in L^2(I), then $f(x) = g(x)$ almost everywhere.

I've twiddled a bit but don't see a way to relate values at a single point to value of the norm. For instance, I start by taking any $x$ such that $f_{n}(x) \rightarrow g(x)$ and any $\varepsilon > 0$ , then there is an $N$ such that $n \geq N \Rightarrow |f_{n}(x) - g(x)| < \varepsilon$ and $||f_{n} - f|| < \varepsilon$ . I can try to square both sides of the last inequality, but it doesn't seem to help. I seem to need to relate the norm of the functions to the value of the functions at x, but I don't see how. I guess I could try to talk about a sequence of upper functions that approaches each function, but that seems like it's way too gritty for this problem. I thought of multiplying the norm times the absolute value, since the absolute value is a constant, but that doesn't seem to help. I'm out of ideas.

I think I sort of solved it a while ago by showing that $||f - g|| = 0$ , but it turns out that proving that this entails the desired consequence requires more measure theory than we've dealt with in our analysis class.

... Well, maybe it would be fruitful to attack it from the definition of "measure 0" which we take to be a set of points which can be covered by neighborhoods, the sum of whose lengths (in $\mathbb{R}$ ) is arbitrarily small. I'm fiddling with it from that angle now.

**Jose27** · August 9th 2010, 06:52 PM

If you want a different appraoch this might interest you (third post first five lines).

**drjerry** · August 13th 2010, 02:27 PM

The problem is subtle. I assume that by "I" you mean $[0,1]$ or some bounded interval.

At first I wanted to suggest using the density of the continuous functions in $L^p$ spaces with a standard $\varepsilon/3$ argument, but that may be too complicated.

Since your domain is compact, $L^2$ -convergence implies $L^1$ -convergence. Do you agree?

Recall that a sequence $(f_n)_n$ converges in measure to $f$ if, for every $\varepsilon > 0$ ,

$\mu(\{ x:|f_n(x) - f(x)| \geq \varepsilon\}) \to 0$ as $n\to 0$ .

The following two intermediate lemmata imply your result.

Lemma 1. If $f_n\to f$ in $L^1$ , then $f_n\to f$ in measure.

Lemma 2. If $f_n \to f$ in measure, then there is a subsequence $(f_{n_k})_k$ such that $f_{n_k} \to f$ a.e.

(I'm paraphrasing from section 2.4 of Folland's book, if you have access to it.)

I haven't looked at the details of Jose's suggestion closely, but my intuition says that the same idea is present there too.

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