# Equivalent operator norm on dense subset

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?

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?.
• 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? – user528502 May 6 '19 at 0:03