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I have the following problem. Show that $y_n={1 \over \sqrt n}\int_0^1t^nx(t)dt$ is a bounded linear operator that maps $L_2[0,1]$ into $l_2$ with the usual norm on the respective spaces.

My approach so far was to look at $\sum_{n=1}^\infty|y_n|^2 =\sum_{n=1}^\infty|{1 \over \sqrt n}\int_0^1t^nx(t)dt|^2\le \sum_{n=1}^\infty{1\over n}\int_0^1t^{2n}|x(t)|^2dt\le \sum_{n=1}^\infty{1\over n}\int_0^1|x(t)|^2dt$ because $t\in [0,1]$. Now the integral exists but the series $\sum_{i=1}^\infty{1\over n}$ diverges and so the product is not finite .

What is the thing to do here?

I hope I've been clear enough, Thanks.

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    $\begingroup$ use the Cauchy-Schwarz inequality on the integral $\int_{0}^{1}|t^n x(t)| dt$ $\endgroup$ – user159517 Jan 24 '15 at 23:00
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From Cauchy-Schwarz inequality we have: $$ y_n^2=\frac{1}{n}\left(\int_{0}^1t^nx(t)\ dt\right)^2\leqslant \frac{1}{n}\int_0^1t^{2n}\ dt\cdot \int_0^1 |x(t)|^2\ dt=\frac{1}{n(2n+1)}\|x\|_2 $$

So, $\|A\|\leqslant \sum_{n=1}^{\infty}\frac{1}{n(2n+1)}<\infty$.

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  • $\begingroup$ All right, thanks. $\endgroup$ – adoion Jan 24 '15 at 23:09

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