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In this question $E$ stands for a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$. Let ${\bf S}=(S_1,\cdots,S_d) \in \mathcal{L}(E)^d$. I want to know if this two assertions are equivalent?

$(1)$ $\|S_kx_n\|\rightarrow \|S_k\|$, as $n\longrightarrow\infty$, for all $1\leq k \leq d$.

$(2)$ $\left(\displaystyle\sum_{k=1}^d\|S_kx_n\|^2\right)^{1/2}\rightarrow \left(\displaystyle\sum_{k=1}^d\|S_k\|^2\right)^{1/2}$, as $n\longrightarrow\infty$.

Thank you for your help.

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  • $\begingroup$ You can forget about all the context and consider sequences of points in $\mathbb{R}^k$. Condition (1) means that the sequence converges point-wise (or component-wise), while condition (2) means that the (euclidean) norm of the point converges to the norm of the limit point. $\endgroup$ – Hellen Oct 16 '17 at 18:15
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If $d > 1$ they are not equivalent.

For $i \in \mathbb{N}$ let $T_i : \ell^2 \to \ell^2$ be a bounded linear map defined as $T_ix = x_i$, for $x = (x_n)_{n=1}^\infty \in \ell^2$, just like in my previous answer. We have established that $\|T_i\| = 1$.

Now define a constant sequence $(x_n)_{n=1}^\infty $ of vectors in $\ell^2$ as $x_n = \sqrt{d}\cdot e_1$, for all $n \in \mathbb{N}$.

We have:

$$\sqrt{\sum_{k=1}^d \|T_kx_n\|^2} = \sqrt{\sum_{k=1}^d d \cdot \delta_{1k}} = \sqrt{d} = \sqrt{\sum_{k=1}^d \|T_k\|^2}$$

However, $\|T_2x_n\| = 0 \not\to 1 = \|T_2\|$. Thus, $(2)$ does not imply $(1)$.

We of course have $(1) \implies (2)$ because of the continuity of $F : \mathbb{R}^d \to \mathbb{R}$ defined as $$F(x_1, \ldots, x_d) = \sqrt{\sum_{k=1}^d |x_k|^2}, \quad (x_1, \ldots, x_d) \in \mathbb{R}^n$$

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  • $\begingroup$ You could actually have chosen $\mathbb R^d$ instead of $\ell^2$. $\endgroup$ – amsmath Oct 16 '17 at 19:10
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If $\|x_n\|\le 1$, then the equivalence is true. It should be clear that (1) implies (2). To go from (2) to (1), assume for simplicity that $d=2$. Put $a_n := \|S_1x_n\|^2$ and $b_n := \|S_2x_n\|^2$. Also, put $a := \|S_1\|^2$ and $b := \|S_2\|^2$. By definition of the operator norm, you know that $a_n\le a$ and $b_n\le b$ for all $n$. Now, you have, $$ 0\le a-a_n = (a+b) - (a_n+b_n) + (b_n-b)\le(a+b) - (a_n+b_n)\to 0. $$ Hence, $a_n\to a$, i.e., $\|S_1x_n\|^2\to\|S_1\|^2$.

Of course, this still holds if $\limsup\|x_n\|\le 1$. But otherwise, there are counterexamples, as you can see in mechanodroid's answer.

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