Equivalence of norms proof

This question is from a set of optional, much harder problems from my first year analysis course, but the subject material is norms on $\mathbb R^K$.

(c) Show that there exists a constant $C > 0$ such that for all $\mathbf x ∈ \mathbb R^k$, $||\mathbf x||\le C||\mathbf x||_\infty$

Hint: Use the standard unit vectors to write $\mathbf x = x_1\mathbf e_1 +. . .+x_n \mathbf e_n$, and use the triangle inequality.

(d) Show that there also exists a constant $c > 0$ such that for all $x ∈ \mathbb R^k$, $||\mathbf x||_\infty \le c||\mathbf x||$

Hint: If it isn’t true, then you can find a sequence $⟨\mathbf x_n⟩$ such that $||\mathbf x_n||$ is bounded but $c_n := ||x_n||_∞ → ∞$. (Why?) Think about the sequence $⟨\mathbf x_n/c_n⟩$. The Bolzano-Weierstrass theorem may be of assistance.

Note that the unspecified norm represents any norm for $\mathbb R^k$. I've done part c), but am stuck on part d). I've had a look for proofs on other questions on MSE, but none of them seem relevant to the hint given, and frequently the terminology is too advanced for me.

I tried considering the case if it isn't true, as suggested, but I can't see how to bound any sequence using that. Any help would be much appreciated.

Solution 1:

Assume that it is not true: $$ \forall c>0 \ \ \exists a_c \ \ \|a_c\|_\infty > c\|a_c\| $$

Note $x_n = a_n / \|a_n\|, c=n\in\Bbb N$ then

$$ \|x_n\| = 1 \\ \|x_n\|_\infty > n \implies \|x_n\|_\infty \to \infty $$

Note $y_n = x_n / \|x_n\|_\infty$ then as $\{ y: \|y\|_\infty = 1 \}$ is compact, you can extract a convergent subsequence (for $\|.\|_\infty$) $y'_n \to y$ and $\|y\|_\infty = 1$.

As $ \|z\| \le C \|z\|_\infty $ you also have $y'_n \to y$ for $\|.\|$, and then $$ \|y'_n\| \to \|y\| $$ But also $$ \|y'_n\| = \frac {\|x_n'\|}{\|x_n'\|_\infty} \to 0 $$

and then $$ \|y\| = 0\implies y= 0 \\ \|y\|_\infty = 1 \implies y\neq 0 $$which is impossible.

Solution 2:

Consider the identity map from $\mathbb{R}^k$ with $|| .||$ to $\mathbb{R}^k$ with $||.||_{\infty}$. By part c) This map is continuous. So by open mapping theorem, the inverse is also continuous, which implies that $||x||_{\infty} \leq ||x||$