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\fancyhead[L]{Hand-in 8 Functional Analysis}
\fancyhead[R]{Thomas van Maaren (9825827)}


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\textbf{Exercise 13} (exercisre similar to Example 7.11). Show that the operator $T:\ell^2\to \ell^2$ defined by
$$(Tx)_n = n^{-3}x_n,\ \ \forall x\in \ell^2$$
is well-defined, bounded and compact
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\textbf{Well-defined:} For any $x\in \ell^2$ we see that

$$\|Tx\|_2^2 = \sum_{n\in \N} n^{-6} |x_n|^2 \leq \sum_{n\in \N}  |x_n|^2 = \|x\|_2^2<+\infty$$

hence $Tx\in \ell^2$.
\textbf{Bounded:} As we saw earlier $\|Tx\|_2 \leq \|x\|_2$ for all $x\in \ell^2$, therefore $\|T\|\leq 1$ and $T$ is bounded.
\textbf{Compact:} For each $k\in \N$ define $T_k:\ell^2\to \ell^2$ by

$$(T_kx)_n = b_n^k,\ \ \text{where }\begin{cases}b_n^k = n^{-3}x_n,&  n\leq k \\ b_n^k=0,&n>k\end{cases}$$

The operators $T_k$ are bounded and linear, and have finite rank. Furthermore for any $x\in \ell^2$ we have

$$\|(T_k - T)x\|_2^2 = \sum_{n=k+1}^\infty |x_n|^2/n^6 \leq (k+1)^{-6}\sum_{n=k+1}^\infty |x_n|^2 \leq (k+1)^{-6} \|x\|_2^2$$

It follows that $$\|T_k - T\| \leq (k+1)^{-1}$$ and so $\|T_k - T\| \to 0$. Thus $T$ is compact by Corollary 7.10.

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