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?
It is true (which seems surprising), being a direct application of the closed graph theorem.
If we have a continuous linear map $T:Y_0\to Y_1$ so that $T(X_0)\subseteq X_1$ denote with $T': X_0\to X_1$ the induced map. It is clearly linear and is a map between Banach spaces. We verify that it is has a closed graph and then by the closed mapping theorem it is continuous.
The condition you need to check is if $x_n\in X_0$ and $T'(x_n)\in X_1$ both converge in $X_0$ and $X_1$ respectively, that then $T'(\lim_n x_n) = \lim T'(x_n)$, where the limits are in $X_0$ and $X_1$ respectively. For this we will use $i_1\circ T' = T\circ i_0$.
So suppose the two limits are not the same, ie $T'(x_n)\overset{X_1}\to y$ and $x_n\overset{X_0}\to x$ with $T'(x)\neq y$. Then it follows that
$$i_1(T'(x)) = (T\circ i_0)(x) = \lim_n (T\circ i_0)(x_n)\neq i_1(y) = i_1(\lim_n T'(x_n)) = \lim_n (i_1\circ T')(x_n)=\lim_n (T\circ i_0)(x_n) $$ which is a contradiction.