# Identity map between two Banach spaces always bounded

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.

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.

• Thank you for your response. Without closed graph theorem, is there a way to show that the identity map is closed?
– Korn
Feb 4, 2021 at 5:16
• @catsadnfish I doubt that you can find an essentially easier way to show this. Feb 4, 2021 at 9:54
• You could also use the Open Mapping Theorem or the Baire Category Theorem. Feb 4, 2021 at 16:48

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 $$\lVert x \rVert_1 = \infty$$.

• But in the OP's case we have the same $X$. That makes a difference. Feb 4, 2021 at 2:39
• That's a fair point!
– JKL
Feb 4, 2021 at 6:29