1
$\begingroup$

$\mathbb{R^n}$ with a norm $||.||$, where $(\mathbb{R^n},||.||)$ is a Banach space.

My question is why the sup-norm $||.||_{\infty}$is equivalent to $||.||$.

My idea:

It suffices to prove that the identity is Continuous: $$I : (\mathbb{R^n},||.||)\rightarrow (\mathbb{R^n},||.||_{\infty})$$ $I(x)=x$, for all $x=(x_1,...,x_n)\in\mathbb{R^n}$.

So, since the two spaces are Banach spaces, with the Closed Graph theorem, it suffices to prove that the graph is closed in the product topology that we define it's norm as follows: $$||(x,y)||_{\mathbb{R^n}\times \mathbb{R^n}}=||x||+||y||_{\infty}$$ So let $(x_n,y_n)\in F$ ,$(F=\{(x,Ix)=(x,x) , x\in\mathbb{R^n}\})$, that converges to $(x,y)$, and we want to prove that $y=Ix=x$.

So we have that for all $n \in \mathbb{N}$ $y_n=Ix_n=x_n.$

and $||x_n-x||\rightarrow 0$

and $||x_n-y||_{\infty}\rightarrow 0$

But how to conclude that $x=y$?

$\endgroup$
1
$\begingroup$

If $(e_i)$ is the usual basis then $\|\sum a_ie_i\| \leq \max \{|a_i|\} \sum \|e_i\|$. Hence $\|x\| \leq C \|x\|_{\infty}$ where $C=\sum \|e_i\|$. It follows from this that you actually have $\|x_n-x\| \to 0$ and $\|x_n-y\| \to 0$. Hence $x=y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.