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Let $X$ and $Y$ be normed vector spaces and let $X_{0}\subset X$ be a dense subspace. Further, let $T:X\longrightarrow Y$ be a bounded, linear operator. Prove that $||T||_{L(X,Y)}:=\underset{x\in X,\\||x||=1}{sup}||Tx||= \underset{x\in X_{0},\\||x||=1}{sup}||Tx||$

The strategy is to prove „$\leq$“ in both directions, whereas $||T||\geq \underset{x\in X_{0},\\||x||=1}{sup}||Tx| $ is clear because $X_{0}\subset X$. However, I do not manage to show the reversed inequality, namely $||T||\leq \underset{x\in X_{0},\\||x||=1}{sup}||Tx||$, because I don‘t see how to use the density property: for all $\epsilon>0$ and for all $x\in X$ there exists $y\in X_{0}$ such that $||x-y||<\epsilon$. Can anyone help me out?

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If $x_0 \in X$ and $\|x_0\|=1$ then there exists $\{x_n\} \subset X_0$ such that $\|x_n-x_0\| \to 0$. Note that $\|x_n\| \to \|x_0\|$. For $n$ sufficiently large $x_n \neq 0$ and $y_n=\frac {x_n} {\|x_n\|}=1$. So $sup \{\|Ty\|: y \in X_0, \|y\|=1\} \geq \|T(y_n)\|$. Can you finish the proof now?.

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  • $\begingroup$ Do you mean $x$ instead of $x_{0}$? By continuity of $T$ we have that for all $\epsilon>0$ there exists $\delta>0$ such that ||y_{n}-x||<\delta implies $||Tx||-||Ty_{n}||\leq ||Ty_{n}-Tx||<\epsilon$. Therefore $||Tx||<||Ty_{n}||+\epsilon$. Since this holds for arbitrary $x$ we can take the sup and let $\epsilon$ go to 0. The result follows. Is it correct? $\endgroup$ – user528502 May 6 '19 at 0:03
  • $\begingroup$ Sorry about the mistype. Yes, your argument is correct. $\endgroup$ – Kavi Rama Murthy May 6 '19 at 0:07
  • $\begingroup$ Thanks for your help! $\endgroup$ – user528502 May 6 '19 at 0:08

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