If $x \in \mathbb{C}^{m}$ and $A$ is a $m \times n$ matrix, prove the following inequalities:
(a) $\|x\|_{\infty} \leq\|x\|_{2}$.
(b) $\|x\|_{2} \leq \sqrt{m}\|x\|_{\infty}$.
(c) $\mid A\left\|_{\infty} \leq \sqrt{n}\right\| A \|_{2}$.
(d) $\|A\|_{2} \leq \sqrt{m}\|A\|_{\infty}$.
How to do this last 2 inequalities I know some how I need to use (a) and (b) and use operator norm but vector product is not defined for this case

My attempt: the language will be $$\mathcal{L} ~\text{is} ~ \{ 0, 1, +, \times, \sin, \cos , \tan , \leq \}$$

The arity of $+$ and $\times$ is two, and that of $\sin, \cos, \tan$ is one. The arity of $\leq$ is two.

Case 1: $a,a^2,\cdots, a^{m-1}$ are all distinct.
Then $a^m \in G$ by closure. It follows that $a^m$ can’t be equal to $a^j$ for any $j: 1\le j\lt m$.

Say we are integrating something as simple as $\frac{x}{e^{x^2}}$ and for some reason you mess up and u-sub. setting $u = e^{x^2}$.
You get something like $\frac{x}{u} \frac{du}{2x e^{x^2}}$ as the integrand. It kind of looks like you could "re sub" for that $e^{x^2}$ in the second fraction to get $\frac{1}{2} \frac{du}{u^2}$ as the integrand, which still integrates to the correct answer. Just wondering if you can "double substitute" like that or if this is some coincidence.