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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.

Masacroso
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    Compact operator is bounded. So you have to show that $\lim_{n\to \infty }|f_n|_2=0$. – D.G Dec 11 '19 at 19:41
  • @Dato we have that $|f_n|_2=\frac1{\sqrt{2+1/n}}$, so it doesn't converge to zero. Indeed the sequence $(f_n)$ doesn't converge in $L^2$ – Masacroso Dec 11 '19 at 19:42
  • Could it be useful to use the characterization of compact operators $T = \sum_{m=1}^\infty \lambda_m \langle u_m, \cdot \rangle v_m$ for suitable (not necessarily complete) orthonormal sets ${u_m}_m$ and ${v_m}_m$? – popoolmica Dec 11 '19 at 19:43
  • @popoolmica this characterization is not known at the time that this exercise appears – Masacroso Dec 11 '19 at 19:47
  • I have edited the answer. In general $\langle Tf_n , v\rangle \to 0$ for all $v$ does not implies that $Tf_n \to 0$ in $L^2$. Compactness of $T$ is needed (Probably you knew, I just make it clear in the answer). – Arctic Char Dec 11 '19 at 21:51

1 Answers1

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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).

Arctic Char
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