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The Daugavet equation is of consequence to the philosophy of linear approximation. Let $E$ be any finite dimensional subspace of $C([0,1])$. Then it follows that there cannot exist any non-zero operator $T:C[0,1]\to E$ such that $T(f)$ is a good approximation of $f$ for all $f\in C[0,1]$. Indeed, because of $\Vert I-T\Vert=1+\Vert T\Vert>1$ we find ...


1

I'll write this since no one has yet written anything ( and, of course, I think this is helpful): An operator between normed spaces is compact if the image of any bounded sequence has a convergent subsequence ( which, if you unravel, is equivalent to other definitions in terms of, e.g, relatively -compact sets ). Showing the right-shift operator in $l^2$ ...


2

Let $T:X\to Y$ be a continous linear operator and $E$ be a bounded set in $X$ Let's call $F = T(E)$, Now the non obvious part is to show that $\overline{F} \subset T(\overline{E})$, this follow from the closed graph theorem Let's assume this property, and take $(y_n)$ a sequence in $\overline{F}$ Now, for each $y_n$, let's take an $x_n \in \overline{E}$ ...


4

Proof by contrapositive certainly works. If $a_{n} \not \to 0$, there is an $\epsilon > 0$ and subsequence $a_{n_k}$ such that $|a_{n_k}| > \epsilon$ for each $k$. Then, show that the sequence $\{e_{n_k}\}$ is mapped to a sequence with no convergent subsequence.


2

That result quoted in the solution is not quite true: what is true is that $\text{Ker } T = (\text{Im } T^*)^{\perp}$. It doesn't matter, though, since $T$ being compact implies that its Hilbert space adjoint $T^*$ is compact as well. Hence, by Theorem 2, $(\text{Im } T^*)^{\perp} \neq \{0\}$, so that there is an $e \neq 0$ in $H$ so that $Te=0$.



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