veneciasp

Answered

2022-07-07

How to solve the following problem:

Let ${f}_{n},f\in {L}^{2}({\mathbb{R}}^{d})$ for all $n\ge 1$ be such that $\Vert {f}_{n}{\Vert}_{2}\to \Vert f{\Vert}_{2}$ as $n\to \mathrm{\infty}$. 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.

Let ${f}_{n},f\in {L}^{2}({\mathbb{R}}^{d})$ for all $n\ge 1$ be such that $\Vert {f}_{n}{\Vert}_{2}\to \Vert f{\Vert}_{2}$ as $n\to \mathrm{\infty}$. 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.

Answer & Explanation

Oliver Shepherd

Expert

2022-07-08Added 24 answers

You want to show

$\Vert f-{f}_{n}{\Vert}_{2}\to 0$

Equivalently, you can show

$\int (f-{f}_{n}{)}^{2}d\mu =\Vert f-{f}_{n}{\Vert}_{2}^{2}\to 0$

We have

$\Vert f-{f}_{n}{\Vert}_{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 $\Vert {f}_{n}{\Vert}_{2}\to \Vert f{\Vert}_{2}$, we have $\int {f}_{n}^{2}d\mu \to \int {f}^{2}d\mu $ and since we have $\int {f}_{n}gd\mu \to \int fgd\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,

$\Vert f-{f}_{n}{\Vert}_{2}^{2}\to 0$

and hence of course also

$\Vert f-{f}_{n}{\Vert}_{2}\to 0$

$\Vert f-{f}_{n}{\Vert}_{2}\to 0$

Equivalently, you can show

$\int (f-{f}_{n}{)}^{2}d\mu =\Vert f-{f}_{n}{\Vert}_{2}^{2}\to 0$

We have

$\Vert f-{f}_{n}{\Vert}_{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 $\Vert {f}_{n}{\Vert}_{2}\to \Vert f{\Vert}_{2}$, we have $\int {f}_{n}^{2}d\mu \to \int {f}^{2}d\mu $ and since we have $\int {f}_{n}gd\mu \to \int fgd\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,

$\Vert f-{f}_{n}{\Vert}_{2}^{2}\to 0$

and hence of course also

$\Vert f-{f}_{n}{\Vert}_{2}\to 0$

bandikizaui

Expert

2022-07-09Added 7 answers

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

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