# Bounded operator mapping onto a subspace

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.

• A bounded operator onto the subspace from where exactly? – Wraith1995 Jun 1 at 1:05
• @Wraith1995 Any Banach space. – Guest Jun 1 at 8:19
• 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/… – plebmatician Jun 1 at 10:38
• @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? – Guest Jun 1 at 10:58

(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 $$V$$.
• @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 $V$. – Jochen Glueck Jun 7 at 19:50