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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?

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  • $\begingroup$ What are $X,Y$? $\endgroup$ Jan 5 '20 at 15:14
  • $\begingroup$ My bad. $T \in L (Y_0, Y_1)$ $\endgroup$ Jan 5 '20 at 15:17
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    $\begingroup$ That seems false to me. Are you sure $X_1$ only has a continuous inclusion into $Y_1$ and not a bi-lipschitz inclusion? (or the norm on $X_1$ is the restriction of the norm on $Y_1$) $\endgroup$
    – s.harp
    Jan 5 '20 at 16:49
  • $\begingroup$ I don't really understand what you mean. $i_0$ and $i_1$ are embeddings, $T$ is a linear and continuous operator from $Y_0$ onto $Y_1$. $\endgroup$ Jan 5 '20 at 17:05
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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.

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  • $\begingroup$ Is $x_n \in X_0 $ ? $\endgroup$ Jan 5 '20 at 19:29
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    $\begingroup$ @s.harp Could you have a look at my follow-up question? math.stackexchange.com/questions/3498621/… I thought that I had a counterexample, but your proof looks very convincing and still I cannot find my mistake. $\endgroup$ Jan 5 '20 at 20:42
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    $\begingroup$ @SeverinSchraven RhysSteele has already pointed out an error in your example. In trying to find counter-examples I made almost exactly the same error (and a whole bunch of other ones too). In fact there are 3 deleted answers to this post describing counter examples, so really you are not alone in believing this statement must be false, only to be frustrated by some detail not working as expected in a counter-example! $\endgroup$
    – s.harp
    Jan 5 '20 at 23:59
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    $\begingroup$ @MathStudent yes, the sequence $x_n$ is in $X_0$, $i_0(x_n)$ is in $Y_0$, $T'(x_n)\in X_1$ and $i_1(T'(x_n))=T(i_0(x_n))\in Y_1$. That can be confusing, since I didn't write explicitly where the sequence must live. $\endgroup$
    – s.harp
    Jan 6 '20 at 0:00

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