# If $\|x_n-x\|_1\to 0$, why $\|Tx_n\|_2\to\|Tx\|_2$?

Let $\|\cdot\|_1$ and $\|\cdot\|_2$ be two norms on a vector space $F$.

Assume that:

• there exists $(x_n)_n\subset F$ such that $\lim_{n\to \infty}\|x_n-x\|_1=0$ .

• $T$ is continuous from $(F,\|\cdot\|_2)$ to $(F,\|\cdot\|_2)$.

I want to prove that $$\lim_{n\to \infty}\|Tx_n\|_2=\|Tx\|_2.$$

I ask this question because I don't find an explication to the following result which I find it in a paper:

Let $F$ stands for a complex Hilbert space with inner product $\langle\cdot\;,\;\cdot\rangle$ and the norm $\|\cdot\|$. Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on $F$.

Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$) and $P$ be the orthogonal projection of $F$ onto the closure of $\text{Im}(M)$.

• $\text{Im}(M^{1/2})$ endow with the inner product $$(M^{1/2}x,M^{1/2}y)_{\text{Im}(M^{1/2})}:=\langle Px, Py\rangle,\;\forall\, x,y \in F,$$ is a Hilbert space.

• For any $x\in F$ there exists a sequence $(x_n)_n$ with $$M^{1/2}x=\lim_{n\to \infty} Mx_n \Longleftrightarrow \lim_{n\to \infty}\|Mx_n-M^{1/2}x\|=0.$$

If $S\in\mathcal{B}(\text{Im}(M^{1/2}))$, why for all $x\in F$, we have $$\|SM^{1/2}x\|_{\text{Im}(M^{1/2})}=\lim_{n\to \infty} \|SMx_n\|_{\text{Im}(M^{1/2})}\;?$$

• @b00nheT I assume that $\|x_n-x\|_1\to 0$ but I don't assume that $\|x_n-x\|_2\to 0$ – Schüler Aug 12 '18 at 9:03
• I think you need to specify the relationship between $\| \cdot \|_1$ and $\| \cdot \|_2$ to be able to prove this. – Theoretical Economist Aug 12 '18 at 9:05
• Well: if you have no information at all on the norms, the statement is false in general. You just need to come up with some smart counterexample – b00n heT Aug 12 '18 at 9:06
• Is your vector space finite dimensional? – user370967 Aug 12 '18 at 9:07
• The statement is false then. I suggest you try $T = 1$ – user370967 Aug 12 '18 at 9:12

The result is false. Consider for example $F=C([0,1])$ with norms $\|\ \|_\infty$ and $\|\ \|_1$ and $T$ the identity operator. It is easy to find a sequence $x_n(t)$ such that $\|x_n\|_1\to0$ and $\|x_n\|_\infty$ does not converge.

Now let's look at the original the problem. We have $Mx=M^{1/2}M^{1/2}x\in\text{Im}(M^{1/2})$. \begin{align} \|Mx_n-M^{1/2}x\|_{\text{Im}(M^{1/2})}&=\|M^{1/2}M^{1/2}x_n-M^{1/2}x\|_{\text{Im}(M^{1/2})}\\ &=\|PM^{1/2}x_n-PM^{1/2}x\|\\ &\le\|M^{1/2}x_n-M^{1/2}x\|, \end{align} and the last expression converges to $0$.