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I'm reading a multivariable calculus textbook for college, and before a Taylor series proof is given, a lemma is provided with its demonstration.

The lemma says: Given a matrix $M(X)= \begin{pmatrix} & A_1\cdot X & \\ & A_2\cdot X \\ & \vdots & \\ & A_n\cdot X & \end{pmatrix}$ such that $A_i\in\mathbb{R}^{n\times n}$ and $X\in\mathbb{R}^n$. Then: $$\|M(X)\|_\infty\le \left(\sum_{i=1}^{n} {\|A_i\|_\infty}^2\right)^{\frac{1}{2}}\|X\|$$

The proof is short. The author simply claims that given $Y\in\mathbb{R}^n, \|Y\|\le1$, then, because $$\|M(X)\cdot Y\|^2=\sum_{i=1}^{n} |(A_i\cdot X)\cdot Y|^2$$ and for each $i$ between $1$ and $n$: $$|(A_i\cdot X)\cdot Y|\le\|A_i\cdot X\|\|Y\|\le\|A_i\cdot X\|\le\|A_i\|_\infty\|X\|$$ Then the proof is done (literally, he doesn't end it). As far as my knowledge goes, that merely implies that $$\|M(X)\cdot Y\|\le\left(\sum_{i=1}^{n} {\|A_i\|_\infty}^2\right)^{\frac{1}{2}}\|X\|$$ But $\|M(X)\cdot Y\|\le\|M(X)\|_\infty$, because $\|Y\|\le1$. That is actually the way that $\|M(X)\|_\infty$ is defined in my textbook. Please tell me if I'm missing something. Thank you very much.

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You have $$\|M(X)\cdot Y\|\le\left(\sum_{i=1}^{n} {\|A_i\|_\infty}^2\right)^{\frac{1}{2}}\|X\|$$

Now observe that this inequality holds for all $Y$ with $\|Y\| \leq 1 $ and that the right side of the inequality is independent of $Y$, which makes it an upper bound for $\|M(X)\cdot Y\|$, therefore it must be greater or equal to the supremum of this expression.

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  • $\begingroup$ Absolutely correct, except you should have written "greater or equal". I really don't know how I missed it, thank you very much! $\endgroup$ – sbs95 Feb 25 '15 at 9:29
  • $\begingroup$ @sbs95 You're of course right. We all miss obvious/easy things sometimes. $\endgroup$ – user159517 Feb 25 '15 at 14:06

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