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Let $g\in L^2(-1,1)$ and let $f_n$ be a sequence of functions in $L^2(-1,1)$. We can use the continuity of the inner product in Hilbert spaces to switch limits and integrals: $$ \int_{-1}^1 \lim_{n\to \infty} f_n(x) g(x) dx = \langle \lim_{n\to\infty}f_n, g \rangle = \lim_{n\to\infty} \langle f_n, g \rangle = \lim_{n\to \infty} \int_{-1}^1 f_n(x) g(x) dx. $$ Instead of using the continuity of the inner product directly, is it possible to show that limit and integral of $\int_{-1}^1 \lim_{n\to \infty} f_n(x) g(x) dx$ can switched using the monotone convergence theorem or the dominated convergence theorem?

uniquesolution
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ManUtdBloke
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  • It miss detail ! Does $f_n$ converge weakly in $L^2$ ? If yes, you can, otherwise, there is no reason. – Surb Aug 31 '17 at 12:28
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    Using convergence theorems directly will not work. For one, we do not know that $f_n\to f$ almost everywhere, we only know that $f_n\to f$ in $L^2$. Even if we do have almost everywhere convergence, there appears to be no assumption of monotonicity, and we cannot necessarily get a dominating function. – Jason Aug 31 '17 at 13:01
  • Why do we need monotonicity, the dominated convergence theorem doesn't assume that requirement, it just requires pointwise convergence? – ManUtdBloke Sep 01 '17 at 06:47

1 Answers1

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Here is one way you could use the dominated convergence theorem if you really wanted to. Assume $f_n$ tends to $f$ a.e. and in $L_2$. Define $S_n=\{x\in\mathbb R \mid |f_n(x)-f(x)|\leq g(x)\}$. Then $$\int_{S_n} (f_n-f) g \to 0$$ by dominated convergence. But $$\int_{\mathbb R\setminus S_n} (f_n-f) g \leq \int (f_n-f)^2 \to 0$$ so $\int f_ng\to \int fg$ as required.

Dap
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