# First isomorphism theorem for Banach Spaces

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?

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

• Why is $\left \|[x]\right \|$ equal to that infimum? May 6, 2018 at 0:04
• @solomeoparedes What is the definition of the norm on $E/\ker{T}$? ;) May 6, 2018 at 0:10
• 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 $\ker T$. May 6, 2018 at 0:15
• @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. May 6, 2018 at 0:47