It is given that $X$ and $Y$ are both Banach spaces. T is an onto bounded linear operator $T:X\rightarrow Y$. I need to show that $Y$ is isomorphic to a quotient space of $X$ and that $Y^{*}$ is isomorphic to a subspace of $X^{*}$.

The notation $X^{*}$ is the dual space.

The idea that I have for the first part is that I can define an equivalence relation that identifies those $x\in X$ that gives the same $y\in Y$, e.g. if $T(x_{1})=T(x_{2})$ then we identify them. Is my idea correct? But how to write it down explicitly and is the isomorphism between $Y$ and the quotient space of $X$ still $T$?

The second part on the dual space I am totally clueless as I am quite terrible at dual spaces. Any help would be greatly appreciated.


Essentially this is the first isomorphism: Define $\widetilde{T}:X/\ker(T)\rightarrow Y$ by $\widetilde{T}([x])=Tx$.

Note that $\widetilde{T}$ is bijective. By the virtue of Open Mapping Theorem, we need only to show that $\widetilde{T}$ is bounded. Indeed, we have \begin{align*} \sup\{\|\widetilde{T}([x])\|: [x]\in U_{X/\ker(T)}\}&=\sup\{\|\widetilde{T}(\pi(x))\|: x\in U_{X}\}\\ &=\sup\{\|Tx\|: x\in U_{X}\}, \end{align*} where $\pi:X\rightarrow X/\ker(T)$, $x\rightarrow[x]$, the canonical map. Here $U_{X/\ker(T)}=\{[x]\in X/\ker(T):\|[x]\|<1\}$ and $U_{X}=\{x\in X: \|x\|<1\}$ and note that $\pi(U_{X})=U_{X/\ker(T)}$.

For the second question, since we have $Y\cong X/\ker(T)$, then $Y^{\ast}\cong(X/\ker(T))^{\ast}$. Now we claim that $(X/\ker(T))^{\ast}\cong\ker(T)^{\perp}$, where \begin{align*} \ker(T)^{\perp}=\{x^{\ast}\in X^{\ast}: x^{\ast}x=0~\text{for each }x\in\ker(T)\}. \end{align*} The mapping is such that $x^{\ast}\in\ker(T)^{\perp}\rightarrow x^{\ast}([x])=x^{\ast}x$.

  • $\begingroup$ Amazing! you made it look so simple haha $\endgroup$ – LanaDR Mar 30 '18 at 7:53
  • $\begingroup$ The U is an open set am I right? And is there a need to show the map at your last line is bounded like the first part to show that it is an isomorphism? $\endgroup$ – LanaDR Mar 30 '18 at 8:01
  • $\begingroup$ The $U$ I have clarified what it is. For the second part, the proof is pretty much similar to the first part. $\endgroup$ – user284331 Mar 30 '18 at 8:02
  • $\begingroup$ Got it man...thanks! I will try to digest it step by step $\endgroup$ – LanaDR Mar 30 '18 at 8:04
  • $\begingroup$ Hi user284331, may I ask you a few questions? 1. $\tilde{T}$ is bijective since the $T$ induces an injective homomorphims for the quotient map and also $T$ is onto (given), am I right? 2. Can you elaborate how the Open Mapping Thm is used in the first part? which step requires the theorem? 3. I don't quite understand the purpose of the "claim" for the second part? Why do you need to show that it is isomorphic to another space $ker(T)^{\bot}$ 4. You used the fact that if there exists an isomorphism between two spaces, then same for their dual spaces. Is there a theorem for this? $\endgroup$ – LanaDR Mar 31 '18 at 10:55

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.