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.

  • 1
    $\begingroup$ 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


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 $V$.

  • $\begingroup$ Can someone explain why V has to be complete? $\endgroup$ – Josh Messing Jun 7 at 17:04
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.