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I have the following proof in my lecture notes:

Fix a sequence $w = (w_n)_{n \geq 1}$ witn $w \in \ell^{\infty}$. Then define $D_w: \ell^2\to \ell^2$ by: $$D_wx = (w_nx_n)_{n \geq 0},$$ Find the $\|D_w\|$

i) $\|D_w\| \le \|w\|_\infty$:

Let $x = (x_n)_{n \geq 1}$. $$\|D_wx\|_2^2 = \sum_{j \geq 1}|(D_wx)_j|^2 = \sum_{j \geq 1}|w_jx_j|^2 \le \sum_{j \geq 1}\|w\|_\infty^2|x_j|^2 = \|w\|^2_\infty \sum_{j \geq 1}|x_j|^2 = \|w\|_\infty^2\|x\|_2^2,$$ so that $\|D_wx\|_2 \le \|w\|_\infty\|x\|_2$. From this we get $\|D_w\| \le \|w\|_\infty$.

ii) $\|D_w\| \ge \|w\|_\infty$:

Since $\|w\|_\infty = \sup\limits_{j \ge 1} |w_j|$, by definition of $\sup$ $\forall \epsilon >0\space \exists j_{\epsilon} \in \mathbb{N} $ such that $|w_{j_{\epsilon}}|\ge \|w\|_{\infty} -\epsilon$

Taking $x=e_{j_{\epsilon}}$ which is the $j_{\epsilon}$-th canonical basis vector ($0$ everywhere except at the $j_{\epsilon}$-th position) , we have that $D_w(e_{j_{\epsilon}})=w_{j_{\epsilon}}e_{j_{\epsilon}}$ and $\|e_{j_{\epsilon}}\|=1$, so

$\forall \epsilon >0:$ $\frac{\|D_w(e_{j_{\epsilon}})\|_2}{\|e_{j_{\epsilon}}\|_2} = |w_{j_{\epsilon}}| > \|w\|_{\infty} -\epsilon$ ....(1)

Then because it is true $\forall \epsilon >0$, making epsilon go to zero:

$\|D_w\|=\sup\limits_{x \neq 0}\frac{\|D_w(x)\|_2}{\|x\|_2} \ge \|w\|_{\infty}$, then $\|D_w\| \ge \|w\|_\infty$ ....(2)

I am having trouble with the second part , I am not convinced how to go from (1) to (2). Since in (1) we are using a particular sequence but in (2) we have $x$. And when taking the limit for $\epsilon \to 0$ in (1) I don't know how the left side transforms to that of (2) Can someone please help filling in the missing details? Thanks.

I found this post "https://math.stackexchange.com/questions/1337896/norm-of-multiplication-operator-in-ell2-bbb-n-is-x-infty" , but is the same as mine and it is missing all the last-part details I am having trouble with

some_math_guy
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1 Answers1

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The norm is defined as the $\sup$ of the norm of the operator applied to all unit vectors. #1 showed that you can find a unit vector to so that you get as close to $\|w_\infty \|$ as you want. Thus the supremum has to be at least $\|w_\infty \|$

Alan
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  • Is there any way to do this in a more systematic way? I don't see how the left-hand side of(1) becomes (2) by taking the limit $ \epsilon \to 0$, since I don't know how this affects the sequence $e_{j_{\epsilon}}$ so that it can become x – some_math_guy Jul 04 '21 at 22:21
  • It doesn't become a particular $x$. What it shows is that there's always a different element $x_{e_\epsilon}$ so that when you apply the operator to it, you get closer and closer to $|w|$ – Alan Jul 04 '21 at 22:23
  • (1) tells me you can get as close as you want by using unit vectors, which are very particular sequences, it doesn't tell me anything about other sequences, so how can the behavior for unit vectors be generalized to any sequence in (2)? – some_math_guy Jul 04 '21 at 22:43
  • (2) is using the non-unit vector version of the norm, note that they divide by the norm of $x$ which turns it into a unit vector. – Alan Jul 04 '21 at 23:18