Given $f\in L^2(\mu)$, $\{f_n\}\in L^2(\mu)$, $f_n\rightarrow f$ a.e., and $\int|f_n|^2\rightarrow\int|f|^2$, then $f_n\rightarrow f$ in $L^2(\mu)$.

Question

I am an undergraduate student taking a second course in real analysis. This is a homework problem.

The problem also suggests using Egorov's theorem. I have made an attempt but I couldn't finish the last step (maybe my approach is wrong).

Here it is:

[Need to show $\int|f_n-f|^2 \rightarrow 0$.]

Let $\epsilon > 0$.

Define $B_m = \{x\in X: d(0,x)<m\}$, for $m\in\mathbb{N}$.

Since $|f|^2$ is integrable there exists $\delta > 0$ such that $\int_E |f|^2 < \frac{\epsilon}{4}$ whenever $\mu(E)<\delta$. (I think this is called "absolute continuity").

By Egorov's theorem there exists $A_\delta\subset E$ such that $f_n\rightarrow f$ uniformly on $A_\delta$ and $\mu(B_m - A_\delta)<\delta$.

Let $E = B_m - A_\delta$.

Then for all $n$

\begin{aligned} \int_{B_m}|f_n-f|^2 &= \int_{A_\delta}|f_n-f|^2 + \int_{E}|f_n-f|^2 \\ &\le \int_{A_\delta}|f_n-f|^2 + \int_{E}|f_n|^2 + \int_{E}|f|^2 \end{aligned}

Now choose an integer $N$ such that for all $n > N$

$$|f_n(x)-f(x)|<\frac{\epsilon}{4\mu(A_\delta)}$$

on $A_\delta$ and

\begin{aligned} &\left|\int_{E}|f_n|^2 - \int_{E}|f|^2\right| < \frac{\epsilon}{4} \\ \Rightarrow &\int_{E}|f_n|^2 < \int_{E}|f|^2 + \frac{\epsilon}{4} \end{aligned}

[Satisfying this second condition is possible because $\int|f_n|^2\rightarrow\int|f|^2<\infty\Rightarrow\int_E|f_n|^2\rightarrow\int_E|f|^2$ (from here).]

But $\int_{E}|f|^2 < \frac{\epsilon}{4}$ by absolute continuity since $\mu(E)<\delta$.

So for all $n > N$ \begin{aligned} \int_{B_m}|f_n-f|^2 &\le \int_{A_\delta}|f_n-f|^2 + \left(\int_{E}|f|^2 + \frac{\epsilon}{4}\right) + \int_{E}|f|^2 \\ &< \int_{A_\delta}|f_n-f|^2 + \left(\frac{\epsilon}{4} + \frac{\epsilon}{4}\right) + \frac{\epsilon}{4} \\ &< \int_{A_\delta}\frac{\epsilon}{4\mu(A_\delta)} + \frac{3\epsilon}{4} \\ &< \frac{\epsilon}{4\mu(A_\delta)}\mu(A_\delta) + \frac{3\epsilon}{4} \\ &=\epsilon \end{aligned}

So the problem is shown if $|f_n - f|^2$ has bounded support (i.e. if the support is contained in some $B_m$)...

[Continuous functions with compact support are dense in $L^p(\mu)$, $1 \le p < \infty$. Does this help me?]

I would really appreciate any hints, direction, or full proof you could provide. (I already submitted the homework.)

Answer 1

Recall that for any inner-product space, the norm induced by the inner-product satisfies the so-called parallelogram identity $\left\Vert a+b\right\Vert ^{2}+\left\Vert a-b\right\Vert ^{2}=2\left\Vert a\right\Vert ^{2}+2\left\Vert b\right\Vert ^{2}$.

Therefore, we have $\left\Vert f_{n}-f\right\Vert ^{2}=2\left\Vert f_{n}\right\Vert ^{2}+2\left\Vert f\right\Vert ^{2}-\left\Vert f_{n}+f\right\Vert ^{2}$. It is given that $||f_{n}||^{2}\rightarrow||f||^{2}$. Moreover, by Fatou lemma, \begin{eqnarray*} 4\int|f|^{2} & = & \int\lim|f_{n}+f|^{2}\\ & \leq & \liminf_{n}\int|f_{n}+f|^{2}\\ & = & \liminf_{n}||f_{n}+f||^{2}. \end{eqnarray*} It follows that \begin{eqnarray*} \limsup_{n}\left\Vert f_{n}-f\right\Vert ^{2} & \leq & \limsup_{n}\left(2\left\Vert f_{n}\right\Vert ^{2}+2\left\Vert f\right\Vert ^{2}\right)+\limsup_{n}\left(-\left\Vert f_{n}+f\right\Vert ^{2}\right)\\ & = & 4||f||^{2}-\liminf_{n}||f_{n}+f||^{2}\\ & \leq & 4||f||^{2}-4||f||^{2}\\ & = & 0. \end{eqnarray*} Therefore $||f_{n}-f||^2\rightarrow0$. (In the above, $||\cdot||$ of course denotes the $L^{2}$-norm)

Answer 2

Put $$A_n = \int ((f_n)_+ -f_+)^2 d\mu = \int (f_n)_+^2 d\mu + \int f_+^2 d\mu -2 \int (f_n)_+ f_+ d\mu,$$ and $$B_n = \int ((f_n)_- -f_-)^2 d\mu = \int (f_n)_-^2 d\mu + \int f_-^2 d\mu -2 \int (f_n)_- f_- d\mu.$$ Hence $$A_n + B_n = \int f_n^2 d\mu + \int f^2 d\mu -2\left(\underbrace{\int (f_n)_+ f_+ d\mu + \int (f_n)_- f_- d\mu}_{C_n}\right).$$ We have $(f_n)+ \to f_+$ and $(f_n)_- \to f_-$ a.e, hence by Fatou lemma, we have $$\liminf C_n \geq \liminf \int (f_n)_+ f_+ d\mu + \liminf \int (f_n)_- f_- d\mu = \int f_+^2 d\mu + \int f_-^2 d\mu = \int f^2 d\mu.$$ In other hand, we have $$C_n \leq \int \frac{(f_n)_+^2 +f_+^2}2 d\mu + \int \frac{(f_n)_-^2 + f_-^2} 2 d\mu = \int\frac{f_n^2 + f^2}2 d\mu, $$ hence $$\limsup C_n \leq \int f^2 d\mu.$$ Thus, we have $C_n \to \int f^2 d\mu$ and then $A_n + B_n \to 0$. Since $A_n, B_n \geq 0$, we must have $A_n \to 0$ and $B_n \to 0$. Using the inequality $(a+b)^2 \leq 2(a^2 + b^2)$, we have $$0\leq \int (f_n -f)^2 d\mu \leq 2(A_n + B_n)$$ which implies $$\lim_{n\to \infty} \int (f_n -f)^2 d\mu = 0.$$