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I am wondering what are the best known relationship between $\|A\|_{2\rightarrow 2}$ and $\|A\|_{\infty\rightarrow 2}$ and how tight it is.

E.g., the trivial result is that for matrix $A$ with dimension $d\times d$

$$\|A\|_{2\rightarrow 2} \le \|A\|_{\infty\rightarrow 2}\le \sqrt{d} \|A\|_{2\rightarrow 2}$$

In particularly I am wondering whether the first inequality is tight (up to universal constant factors), and if yes, do we have a good understanding of when it's close to be tight? (e.g., do we have any existing construction of $A$ so that it's tight?)

(I was asking about the second inequality but I realized that I meant the first inequality)

The same question can be asked for other induced norms as well, which I am also curious about.

Thanks!

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    $\begingroup$ Yes, it is tight: take $A=I$, for example. $\endgroup$
    – Alexandre Eremenko
    Commented Mar 19, 2019 at 11:57
  • $\begingroup$ sorry I meant whether the left-hand side is tight up to universal constant? Is that also obvious? $\endgroup$
    – Alex Wenxin Xu
    Commented Mar 21, 2019 at 0:50
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    $\begingroup$ Yes, it's obvious. Take $A$ to be the matrix with a $1$ in the $(1,1)$ entry and $0$'s everywhere else. $\endgroup$
    – Nik Weaver
    Commented Mar 21, 2019 at 3:58

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It is more a question on norms on spaces and not on operators as $\| \cdot \|_{N_1, N_3} \leq C \| \cdot \|_{N_2,N_3}$ if and only if $N_2 \leq \frac{1}{C} N_1$.

Doing so if $l_{x,y}(z) = \left< x, z\right>y$ then $\| l_{x,y} \|_{\infty, 2} = \|x\|_1 \|y\|_2$ and$\| l_{x,y} \|_{2, 2} = \|x\|_2 \|y\|_2$

And for $x= (1, \cdots,1)$, you have your equality in your inequality.

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  • $\begingroup$ We can think about the behavior of $L^p \mapsto L^q$ operators, where $p = 2= q$ and $p = + \infty, q = 2$. For example $T(f) = f \varphi$ for a suitable $\varphi$. Thus, we can apply a interpolation theorem to compare the norms of theses operators. For example, the Riesz-Thorin theorem: en.wikipedia.org/wiki/Riesz%E2%80%93Thorin_theorem $\endgroup$
    – Marcelo Ng
    Commented Mar 19, 2019 at 12:46
  • $\begingroup$ Sorry I meant to ask if the first inequality is tight $\endgroup$
    – Alex Wenxin Xu
    Commented Mar 21, 2019 at 0:54
  • $\begingroup$ Then take $x = (1, 0, \cdots, 0)$. The point is that your question is just on comparaison of classical norms in a finite dimensional real vector space. $\endgroup$
    – InfiniteLooper
    Commented Mar 21, 2019 at 9:10
  • $\begingroup$ And by such not of research level. See MathSE for questions of this type. $\endgroup$
    – InfiniteLooper
    Commented Mar 21, 2019 at 9:11

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