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Consider a vector $y\in \mathbb{R}^n$, which is defined by $$ y = Wx $$ where $x\in \mathbb{R}^n$ and $W\in \mathbb R^{n\times n}$. I want to induce the following inequality: $$ \|y\|_2\leq \|W\|\|x\|_2 $$ where $\|\cdot\|_2$ is the $2$-norm. In this case, I thought that we should use the spectral norm for $\|W\|$, which is the largest singular value of $W$ because it is the corresponding matrix norm induced by the vector $2$-Euclidean norm.

My question is that I can safely replace the spectrom norm for $W$ with the Frobenius norm? That is, does the following inequality make sense? $$ \|y\|_2\leq \|W\|_F\|x\|_2 $$ It seems that the above inequality is used in the paper I am reading.

user155214
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    The induced norm on operators: $|W|=\sup_{x\neq0}\frac{|Wx|_2}{|x|_2}$. – Anthony Jul 21 '22 at 11:10
  • A minor correction. The $l_{2}$ norm should be the 2-Euclidean norm! $l_{2}$ is an infinite dimensional space, whereas you are working in $\mathbb{R}^{n}$. I know that this term is also used but I think it only confuses people! –  Jul 21 '22 at 11:24
  • @AnthonySaint-Criq Thank you for your comment. The definition you write is equivalent to the largest singular value of $W$. – user155214 Jul 21 '22 at 11:27
  • @GeorgeTsoutsinos Thanks for your correction! – user155214 Jul 21 '22 at 11:27

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You can immediately prove that $\lVert Wx\rVert_2\le \lVert W\rVert_F \lVert x\rVert_2$ by the Cauchy-Schwarz inequality in the form $$ (\sum_j a_jb_j )^2 \le \sum_{j_1}a_{j_1}^2 \sum_{j_2}b_{j_2}^2.$$ Indeed, $$ \lVert Wx\rVert_2^2=\sum_i (\sum_j W_{i{\,}j}x_j)^2 \le \sum_i \sum_{j_1} W_{i{\,}j_1}^2\sum_{j_2} x_{j_2}^2 =\lVert W\rVert_F^2\lVert x\rVert_2^2.$$

Giuseppe Negro
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