# A map between Banach spaces is continuous - counterexample

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.