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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.

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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.

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  • $\begingroup$ Thank you for your response. Without closed graph theorem, is there a way to show that the identity map is closed? $\endgroup$
    – Korn
    Feb 4 at 5:16
  • $\begingroup$ @catsadnfish I doubt that you can find an essentially easier way to show this. $\endgroup$ Feb 4 at 9:54
  • $\begingroup$ You could also use the Open Mapping Theorem or the Baire Category Theorem. $\endgroup$ Feb 4 at 16:48
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Edit: this answer only makes sense if the $X$ is different; see Robert Israel's answer instead.


Given your assumption $\lVert x \rVert_1 \leq M \lVert x \rVert_2$ for some $M > 0$, we can only say that the identity map $I: (X, \lVert \cdot \rVert_2) \to (X, \lVert \cdot \rVert_1)$ is continuous.

To see this, note that if $\lVert x \rVert_2 < \epsilon / M$, then $\lVert I(x) \rVert_1 = \lVert x \rVert_1 \leq M \lVert x \rVert_2 < \epsilon$.

The fact that $X$ associated with either metric being a Banach space doesn't really matter here; this is a more general result from metric/topological spaces. We see that the topology induced by the norm $\lVert \cdot \rVert_1$ is finer than the one induced by the norm $\lVert \cdot \rVert_2$.

If the two norms were equivalent, i.e. $m \lVert x \rVert_2 \leq \lVert x \rVert_1 \leq M \lVert x \rVert_2$, then their topologies would be the same, and so the identity map would be continuous.

For a counterexample to the original direction, we can consider the Banach spaces $\ell^1$ and $\ell^\infty$ (where $\ell^1 \subseteq \ell^\infty$ or equivalently $\Vert \cdot \rVert_\infty \leq \lVert \cdot \rVert_1$). Consider the identity map $I: \ell^\infty \to \ell^1$, and the element $x = (1, 1, 1, \dots)$ which has $\Vert x \rVert_\infty = 1$ but $\lVert x \rVert_1 = \infty$.

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    $\begingroup$ But in the OP's case we have the same $X$. That makes a difference. $\endgroup$ Feb 4 at 2:39
  • $\begingroup$ That's a fair point! $\endgroup$
    – JKL
    Feb 4 at 6:29

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