Today the following question came up (A map between Banach spaces is continuous):

I'm trying to prove this statement:

Let $(X_0, \| \cdot \|_{X_0})$ and $(X_1, \|\cdot \|_{X_1})$ be Banach spaces and $(Y_0, \| \cdot \|_{Y_0})$ and $(Y_1, \|\cdot \|_{Y_1})$ normed spaces so that $X_0$ is a vector subspace of $Y_0$ and $X_1$ is a vector subspace of $Y_1$.

Further assume that $i_0: (X_0, \|\cdot\|_{X_0}) \rightarrow (Y_0, \|\cdot\|_{Y_0}),\; x \mapsto x$ and $i_1: (X_1, \|\cdot\|_{X_1}) \rightarrow (Y_1, \|\cdot\|_{Y_1}),\; x \mapsto x$ are continuous.

If $T \in L(Y_0, Y_1)$ so that $T(X_0) \subseteq X_1$, define $S: (X_0, \|\cdot\|_{X_0}) \rightarrow (X_1, \|\cdot\|_{X_1}), \;x \mapsto Tx$ and show that $S$ is continous.

Any ideas how to prove it?

I was thinking about it for a moment and thought that I came up with a counterexample. Before I could post it, there was an answer showing that it is indeed true. The proof is rather short and looks very convincing. My question is therefore:

What is wrong with my counterexample?

For a counterexample pick a Banach space $(Z, \Vert \cdot \Vert_Z)$, a discontinuous linear map $C: (Z, \Vert \cdot \Vert_Z)\rightarrow (Z, \Vert \cdot \Vert_Z)$. We define $X:=Z\oplus Z$ with the norm $\Vert (z_1, z_2)\Vert := \Vert z_1\Vert_Z +\Vert z_2\Vert_Z$. Then define two discontinuous linear maps $$A: (X, \Vert \cdot \Vert) \rightarrow (X, \Vert \cdot \Vert), A(z_1, z_2):= C(z_1)+z_2$$ and $$ B: (X, \Vert \cdot \Vert) \rightarrow (X, \Vert \cdot \Vert), B(z_1, z_2):=z_1+ C(z_2). $$ Then we define two new norms on $X$. Namely, we define for $x\in X$ $$ \Vert x\Vert_A := \Vert x\Vert + \Vert Ax\Vert$$ and $$ \Vert x\Vert_B := \Vert x \Vert + \Vert Bx\Vert.$$ Now pick $$(X_0, \Vert \cdot \Vert_{X_0})= (X, \Vert \cdot \Vert_A ) = (Y_0, \Vert \cdot \Vert_{Y_0})$$ and $$ (X_1, \Vert \cdot \Vert_{X_1}) = (X, \Vert \cdot \Vert_B)$$ and $$ (X, \Vert \cdot\Vert) = (Y_1, \Vert \cdot \Vert).$$ We have $$ \Vert i_0 x\Vert_{Y_0} = \Vert x\Vert_{Y_0} =\Vert x\Vert_A = \Vert x\Vert_{X_0}$$ and $$ \Vert i_1 x\Vert_{Y_1} = \Vert x\Vert \leq \Vert x\Vert + \Vert Bx\Vert = \Vert x\Vert_{X_1}$$ Thus, $i_0$ and $i_1$ are continuous. Furthermore, we set $$ T: (Y_0, \Vert \cdot\Vert_{Y_0}) \rightarrow (Y_1, \Vert \cdot \Vert_{Y_1}), x\mapsto Ax.$$ We compute $$\Vert T x\Vert_{Y_1} = \Vert Ax \Vert \leq \Vert x\Vert_{Y_0}$$ Hence, also $T\in L(Y_0, Y_1)$.

You claim now that the map $S: (X_0, \Vert \cdot \Vert_{X_0}) \rightarrow (X_1, \Vert \cdot \Vert_{X_1}), x\mapsto Ax$ is continuous as well. This is not true. Note that $i: Z \rightarrow X, z \mapsto (0,z)$ is continuous. If $S$ was continuous, then also the map $F = S\circ i: (Z, \Vert \cdot \Vert_Z) \rightarrow (X_1, \Vert \cdot\Vert), z \mapsto A(0, z)$ was continuous. This would imply $$ \Vert z \Vert_Z + \Vert C(z) \Vert_Z = \Vert (0,z) \Vert + \Vert B(0,z) \Vert = \Vert (0,z) \Vert_{X_1} = \Vert A(0,z)\Vert = \Vert F(z) \Vert \leq \Vert F \Vert_{op} \Vert z \Vert_Z $$ which tells us that $C$ is continuous, which is a contradiction.


$X$ is not a Banach space with e.g. $\|\cdot\|_A$. If it were we would have that $\|\cdot\|_A$ is equivalent to the usual norm on $X$ by Banach's isomorphism theorem since it is clear that $\|x\| \leq \|x\|_A$. If this is true then there is $c$ such that $\|x\|_A = \|x\| + \|Ax\| \leq c\|x\|$ which implies that $A$ is bounded with norm at most $c-1$. Since the projection maps are continuous, this would imply that $(z_1, z_2) \mapsto C(z_1)$ is continuous which in turn implies that $C$ is continuous.

  • $\begingroup$ Thank you. That was incredibly stupid of me... $\endgroup$ Jan 5 '20 at 21:02
  • 1
    $\begingroup$ It's easy to make mistakes like this one when trying to come up with a counterexample. I spent some time thinking about that question earlier and even posted and then deleted an incorrect counterexample with a worse error than this. $\endgroup$ Jan 5 '20 at 21:04

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