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How to solve the following problem:

Let $f_n, f \in L^2(\mathbb{R}^d)$ for all $n \geq 1$ be such that $\|f_n\|_2 \to \|f\|_2$ as $n \to \infty$. Suppose, moreover, that $$\int f_{n}g \to \int fg $$ for all $g \in L^2(\mathbb{R}^d)$. Then $f_n$ converges to $f$ in $L^2$-norm.

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Try using the parallelogram law. –  t.b. Aug 13 '12 at 11:06

2 Answers 2

You want to show $$ \| f - f_n \|_2 \to 0$$

Equivalently, you can show $$ \int (f-f_n)^2 d \mu = \| f - f_n \|_2^2 \to 0$$

We have $$ \| f - f_n \|_2^2 = \int (f-f_n)^2 d \mu = \int f^2 d \mu - 2 \int f f_n d \mu + \int f_n^2 d \mu$$

Since we have $\|f_n\|_2 \to \|f\|_2$, we have $\int f_n^2 d \mu \to \int f^2 d \mu$ and since we have $\int f_n g d \mu \to \int fg d \mu$ we have $\int f f_n d \mu \to \int f^2 d \mu$ so that

$$ \int f^2 d \mu - 2 \int f f_n d \mu + \int f_n^2 d \mu \to \int f^2 d \mu - 2 \int f^2 d \mu + \int f^2 d \mu = 0$$

that is, $$\| f - f_n \|_2^2 \to 0$$

and hence of course also

$$\| f - f_n \|_2 \to 0$$

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Since $L^2$ is a Hilbert space, you can use the parallelogram identity, as suggested in a comment. More generally, you can also use a property of any uniformly convex Banach space, which is listed as the second property here. A very nice proof appears in Brezis' book on functional analysis.

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