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Let $X$ and $Y$ be Banach space and $W\subseteq Y$ be a subspace. Let $T_1:X\to Y$ be bounded and linear such that $T_1(X)=W$ and $T_2:Y\to Y$ be bounded.

Does there exist a Banach space $Z$ and a surjective bounded operator $T:Z\to\{y\in Y: T_2y\in W\}$?

My attempt: If $T_2$ is bijective, then I know the answer is yes. I couldn't answer the general case.

I considered example of shift and projection operators on $l^\infty$ and could always find a bounded operator but I couldn't generalize it.

Edit: I think I can even handle the case when range of $T_2$ is closed, but I still can't handle the general case.

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    $\begingroup$ A bounded operator onto the subspace from where exactly? $\endgroup$ – Wraith1995 Jun 1 at 1:05
  • $\begingroup$ @Wraith1995 Any Banach space. $\endgroup$ – Guest Jun 1 at 8:19
  • $\begingroup$ I think you need to supply the Banach space from which your operator sends for this to be meaningful.. If the Banach space can be $Y$, then if your subspace is closed (which happens if for instance $T_1$ is open) the projection from $Y$ to your subspace is bounded iff your subspace is complemented. See math.stackexchange.com/questions/465016/… $\endgroup$ – plebmatician Jun 1 at 10:38
  • $\begingroup$ @plebmatician It can be any Banach space. My question just means there exists a Banach space such that this holds. Do you mean it is not true in general? $\endgroup$ – Guest Jun 1 at 10:58
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Yes.

This can be seen as follows:

(1) There exists a complete norm $\|\cdot\|_W$ on $W$ such that the embedding $W \hookrightarrow Y$ is continuous. To see this, use that $T_1$ induces a linear bijection $\tilde T_1: X/\ker T_1 \to W$, which is continuous when $W$ is endowed with the norm from $Y$. Now transport the norm from $X$ to $W$ via the bijection $\tilde T_1$ and denote this new norm on $W$ by $\|\cdot\|_W$; clearly, $W$ becomes complete with this new norm, and the embedding $W \hookrightarrow Y$ is continuous.

(2) Let's use the notation $V := T_2^{-1}(W)$. Now we introduce a new norm $\|\cdot\|_V$ on $V$ by defining $$ \|v\|_V = \|v\|_Y + \|T_2v\|_W. $$ Again, it is not difficult to check that the norm $\|\cdot\|_V$ on $V$ is complete. Hence, we can choose the desired Banach space $Z$ as $(V, \|\cdot\|_V)$ and the desired operator $$ T: Z = (V, \|\cdot\|_V) \to (V, \|\cdot\|_Y) $$ as the identity operator on $V$.

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  • $\begingroup$ Can someone explain why V has to be complete? $\endgroup$ – Josh Messing Jun 7 at 17:04
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    $\begingroup$ @JoshMessing: If $(v_n)$ is a Cauchy sequence in $V$, then it is also a Cauchy sequence in $Y$ and $(T_2v_n)$ is a Cauchy sequence in $W$. This can be used to show convergence of $(v_n)$ in $V$. $\endgroup$ – Jochen Glueck Jun 7 at 19:50

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