Im stuck with this exercise:
Let $T$ a compact operator in $L^2([0,1])$, show that $\lim_{n\to \infty }\|Tf_n\|_2=0$ for $f_n(x):=\sqrt{n}x^n$.
I know that $(f_n)$ is linearly independent and that if $(e_n)$ is an orthonormal sequence then $\lim_{n\to \infty }Te_n=0$. Thus I tried to connect the sequence $(f_n)$ with it orthonormalization using the Gram-Schmidt process but I dont find something useful.
Another way would be show that
$$
\lim_{n\to \infty }\langle Tf_n,v \rangle=\lim_{n\to \infty }\langle f_n,T^*v\rangle=0
$$
for all $v\in L^2([0,1])$. But I dont see how I can show that.
Some help will be appreciated.
Note: in any case I cannot use the fact that the span of $(f_n)$ is dense in $L^2([0,1])$ because this is not presented in the book where this exercise comes from.
Best Answer
Note for any continuous function $g$,
$$\tag{1} | \langle f_n, g\rangle | =\left| \int_0^1 \sqrt n x^n g(x) dx\right| \le \|g\|_\infty \int_0^1 \sqrt n x^n = \| g\|_{\infty } \frac{\sqrt n}{n+1} \to 0$$
as $n\to \infty$. Now for any $h\in L^2$ and for all $\epsilon>0$, there is a continuous function $g$ so that $\| h-g\|_{L^2} <\epsilon$. Thus
\begin{align*} |\langle f_n , h\rangle| &\le |\langle f_n, h-g\rangle| + |\langle f_n, g\rangle|\\ &\le \sup \| f_n\|_{L^2} \| h-g\|_{L^2} + |\langle f_n, g\rangle| \end{align*}
(Note that $\sup \| f_n\|_{L^2}$ is finite). By (1), there is $N\in \mathbb N$ so that $$|\langle f_n , h\rangle| \le (\sup \|f_n\|_{L^2} +1 )\epsilon, \ \ \ \forall n\ge N.$$
Thus
$$\lim_{n\to \infty} \langle f_n, h\rangle = 0, \ \ \ \forall h\in L^2[0,1]$$
This implies that $\{ f_n\}$ converges weakly to $0$ in $L^2$. Since $T$ is compact, this implies that $\{Tf_n\}$ converges strongly to $T 0 = 0$ in $L^2$ (see here).