Let $X$ be a Banach spaces, $\| \cdot \|_1$ and $\| \cdot \|_2$ be two different norms.
Suppose that $\| x\|_1 \leq M\| x\|_2$ for all $x\in X$.
I've seen many places that say that the identity map $I: (X, \| \cdot \|_1) \rightarrow (X, \|\cdot \|_2)$ is always bounded, i.e. continuous.
But I'm not sure why it's true. And where does it use the completeness?
This is how I have attempted to prove it.
Let $x\in X$.
Then $\|I(x) \|_1= \|x \| _2 \leq \| I\|_1 \| x\|_1 \leq M\| x\|_2$.
Not sure how to proceed.
Thank you.
Best Answer
The identity map from $(X, \|\cdot\|_1)$ to $(X, \|\cdot\|_2)$ has a closed graph (because its inverse is continuous), so the Closed Graph Theorem says it is continuous.