prove that $g\geq f^2$

The problem is this:

Let $(f_n)$ a sequence in $L^2(\mathbb R)$ and let $f\in L^2(\mathbb R)$ and $g\in L^1(\mathbb R)$. Suppose that $$f_n \rightharpoonup f\;\text{ weakly in } L^2(\mathbb R)$$ and $$f_n^2 \rightharpoonup g\;\text{ weakly in } L^1(\mathbb R).$$ Show that $$f^2\leq g$$ almost everywhere on $\mathbb R$.

I admit I'm having problems with this since It's quite a long time I don't deal with these kind of problems. Even an hint is welcomed. Thank you very much.

• I haven't tried the problem, but you might want to first try to prove that "weak convergence implies boundedness" – M Turgeon Feb 28 '12 at 0:48
• but that follows by the Banach Steinhouse theorem if i remember well. Simply apply it wrt to the family of evaluation functionals right? – uforoboa Feb 28 '12 at 1:20

Here is another solution. The main idea is to obtain an integral estimate, and use an approximation to the identity to reduce it to a pointwise estimate.

Let $E$ be the set of points $x$ such that $x$ is a Lebesgue point of both $f$ and $g$. Since both $f$ and $g$ are locally integrable, $\mathbb{R}-E$ has measure zero.

Now choose $x\in E$ and consider a cube $C_x(r)$ of side length $r$ centered at $x$. Then

$$\phi_{\epsilon}=\frac{1}{|C_{x}(\epsilon)|}\chi_{C_{x}(\epsilon)}$$

lies in $L^2(\mathbb{R})\cap L^{\infty}(\mathbb{R})$, so

$$(f_n,\phi_{\epsilon})\to(f,\phi_{\epsilon})\quad\text{and}\quad(f_n^2,\phi_{\epsilon})\to(g,\phi_{\epsilon}).$$

By Cauchy-Schwarz inequality applied to $f_n\sqrt{\phi_{\epsilon}}\cdot\sqrt{\phi_{\epsilon}}$,

$$(f_n,\phi_{\epsilon})^2=\left( \int f_n\phi_{\epsilon}\right)^2\leq\int f_n^2\phi_{\epsilon}=(f_n^2,\phi_{\epsilon}).$$

Thus taking $n\to\infty$ we have

$$(f,\phi_{\epsilon})^2\leq(g,\phi_{\epsilon}).$$

Now taking $\epsilon\to 0$ gives $f^2(x)\leq g(x)$, completing the proof.

P.S. I saw a same question on AoPS. I hope that my answer is not a duplicate.

Well it is a property of the weak convergence that every weak convergent sequence is bounded and

$$||f||\leq \lim\inf ||f_{n}||$$

then for every $\Omega \in \mathbb{R}^n$ measurable with finite measure we have

$$\left(\int_\Omega f^2\right)^{\frac{1}{2}}\leq \lim\inf\left(\int_\Omega f_n^2\right)^{\frac{1}{2}}$$

That implies

$$\int_\Omega f^2\leq \lim\inf\int_\Omega f_n^2\tag{1}$$

Note that the function $h$ constant equal to 1 in $\Omega$ and 0 case contrary is in $L^{\infty}{(\mathbb{R})}$ which is the dual of $L^{1}{(\mathbb{R})}$ then

$$\int_{\Omega}f_n^2\rightarrow\int_\Omega g\tag{2}$$

Jointing (1) and (2) we get

$$\int_{\Omega}f^2\leq\int_\Omega g$$

Since $\Omega$ is arbitrary we have $f^2\leq g$ almost everywhere.$\blacksquare$