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In this question $\mathcal{H}$ 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 $\mathcal{B}(\mathcal{H})$ the algebra of all bounded linear operators from $\mathcal{H}$ to $\mathcal{H}$.

Let ${\bf T} = (T_1,...,T_d) \in \mathcal{B}(\mathcal{H})^d$. $\|{\bf T}\|$ is given by \begin{eqnarray*} \|{\bf T}\| &=&\sup\left\{\bigg(\displaystyle\sum_{k=1}^d\|T_kx\|^2\bigg)^{\frac{1}{2}},\;x\in \mathcal{H},\;\|x\|=1\;\right\}. \end{eqnarray*} It is well known that for each $k\in\{1,\cdots,d\}$, we have \begin{eqnarray*} \|T_k\| &=&\sup\left\{\|T_kx\|,\;x\in \mathcal{H},\;\|x\|=1\;\right\}. \end{eqnarray*}

It is always true that $$\|{\bf T}\|=\bigg(\displaystyle\sum_{k=1}^d\|T_k\|^2\bigg)^{1/2}\;??$$

Schüler
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Student
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1 Answers1

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The statement does not hold in general for $d > 1$:

Let $(e_n)_{n=1}^\infty$ be the canonical basis for $\ell^2$. For $i \in \{1, \ldots, d\}$ define $T_i : \ell^2 \to \ell^2$ as $T_ix = x_ie_i$, for every $x = (x_n)_{n=1}^\infty \in \ell^2$.

$T_i$ are bounded:

$$\|T_ix\|_2 = \|x_ie_i\|_2 = \left|x_i\right| \le \|x_i\|$$

So $\|T_i\| \le 1$.

Furthermore, for $x = e_i$ we have

$$\|T_i\| \ge \frac{\|Te_i\|_2}{\|e_i\|_2} = 1$$

Thus, $\|T_i\| = 1$.

Define $\mathbf{T} = (T_1, \ldots, T_n) \in \mathcal{B}\left(\ell^2\right)^d$.

We have:

$$\|\mathbf{T}\| = \sup_{\|x\|_2 = 1} \sqrt{\sum_{k=1}^d\|T_kx\|_2^2} = \sup_{\|x\|_2 = 1} \sqrt{\sum_{k=1}^d\left|x_k\right|^2} \le \sup_{\|x\|_2 = 1} \sqrt{\sum_{k=1}^\infty\left|x_k\right|^2} = \sup_{\|x\|_2 = 1} \|x\|_2 = 1$$

However,

$$\sqrt{\sum_{k=1}^d\|T_k\|^2} = \sqrt{\sum_{k=1}^d 1} = \sqrt{d} > 1$$

Thus, for $d > 1$ we cannot have $\|\mathbf{T}\| = \sqrt{\sum_{k=1}^d\|T_k\|^2}$.

Student
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mechanodroid
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  • @Student Here you go, I just used the same $T_i$ as defined here. What kind of relationship between $\omega(\mathbf{T})$ and $\omega(T_i)$ are you expecting? A closed formula or just some inequalities? – mechanodroid Sep 27 '17 at 19:11
  • I find a contradiction between your answer and the two following answers:https://math.stackexchange.com/questions/4551/how-can-i-prove-supab-sup-a-sup-b-if-ab-ab-mid-a-in-a-b-in-b ; https://math.stackexchange.com/questions/2473122/properties-of-the-norm-of-operator/2473139#2473139 ; could you please explain to me my wrong. Thank you – Student Oct 15 '17 at 10:05
  • According to the two answers, I think that the equality is true but your example shows that we cannot have the equality. – Student Oct 15 '17 at 10:07
  • @Student It is true that for $A, B \subseteq \langle 0, +\infty \rangle$ we have $\sup(A + B) = \sup A + \sup B$ but it is not in general true that for two functions $f, g : X \to \langle 0, +\infty\rangle$ holds $$\sup_{x\in X} \big(f(x) + g(x)\big) = \sup_{x\in X} f(x) + \sup_{x\in X} g(x)$$

    Take $\sin$ and $\cos$ on $[0, \frac{\pi}2]$. We have $\sup_\limits{x\in X} \sin x = \sup_\limits{x\in X} \cos x = 1$, but $\sup_\limits{x\in X} \big(\sin x + \cos x\big) = \sqrt{2}$. The problem is that maxima are achieved at different points. The statement from your question is of this type.

     – mechanodroid Oct 15 '17 at 10:21