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I am dealing with the following exercise:

Let $E$, $F$ be Banach Spaces and $T:E\to F$ a continuous and surjective linear function. Define $\widehat{T}:E/\ker T\to F$ as $\widehat{T}([x])=T(x)$. Prove that $\widehat{T}$ is an isomorphism such that $\left \|T\right \|=\left \|\widehat{T}\right \|$.

First of all, since $T$ is continuous then $\ker T$ is closed, so $E/\ker T$ is a Banach space (because $E$ is a Banach Space). Take $Q:E\to E/\ker T$ the projection. Therefore $\widehat{T}\circ Q=T$, so $\widehat{T}$ is a continuous linear function which is bijective. By the open mapping theorem, it is also an homeomorphism. Moreover, $\left \|T\right \|=\left \|\widehat{T}\circ Q\right \|\leq \left \|\widehat{T}\right \|\left \|Q\right \|\leq\left \|\widehat{T}\right \|$ since $\left \|Q\right \|\leq 1$.

The only thing I need to prove in order to finish the exercise is the other inequality: $\left \|\widehat{T}\right \|\leq \left \|T\right \|$. I do not know how to achieve it. How would you solve this last step?

solomeo paredes
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1 Answers1

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Pick some class $[x] \in E/ker(T)$ and an arbitrary $y \in [x]$. Then $$ \| \hat{T}([x]) \|= \|T(y)\| \leq \|T \| \| y \| $$

Therefore, $$\| \hat{T}([x]) \| \leq \|T \| \inf_{y \in [x]} \{ \|y \| \} =\|T \| \bigl\| [x ]\bigr\|$$

Added If you want to avoid the definition with inf, here is how you can re-write the proof:

Let $x \in E$ and $z \in \ker(T)$. Then

$$ \| \hat{T}([x]) \|= \|T(x-z)\| \leq \|T \| \| x-z \| = \| T \| d(x,z) $$

Therefore $$ \| \hat{T}([x]) \| \leq \| T \| \inf_{z \in \ker(T)} \{ d(x,z) \}= \|T \| \| [x]\| $$

N. S.
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  • Why is $\left |[x]\right |$ equal to that infimum? – solomeo paredes May 06 '18 at 00:04
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    @solomeoparedes What is the definition of the norm on $E/\ker{T}$? ;) – N. S. May 06 '18 at 00:10
  • Maybe I am a little bit confused, but the definition our professor gave us is: the norm in $E/S$ where $S$ is a closed subspace is $\left |[x]\right |=d\left (x,S\right )$. In our case, the norm of $\left |[x]\right |$ is the distance between $x$ and the subspace $\ker T$. – solomeo paredes May 06 '18 at 00:15
  • @solomeoparedes $$\left |[x]\right |=d\left (x,S\right ) = \inf { | x-z | : z \in S }\stackrel{y=x-z}{=} \inf { | y| : y \in [x] }$$ I am used with the definition $$| [x] | = \inf_{y \in [x]} { |y | }$$ but as the above line shows they are equivalent. – N. S. May 06 '18 at 00:47