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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.

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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 $x^{\ast}\in\ker(T)^{\perp}\rightarrow x^{\ast}([x])=x^{\ast}x$.

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  • $\begingroup$ 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

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