Show that the following linear transformations are isometric. 
Let $X$ be a Banach space. Let $L\subset X$ is a closed linear manifold. $P:X\rightarrow X/L$ is the quotient map. Show that the following linear transformations are isometric.
  
  
*
  
*$\gamma:(X/L)^*\rightarrow L^a$ where $\gamma(\phi)=\phi\circ P$. 
  
*$\beta: X^*/L^a\rightarrow L^*$ where $\beta(\phi+L^a)=\phi|_L$
  
  
  (Here $L^a=\{\phi \in X^*: \phi(x)=0 \;\text{for all} \;x\in L\}$)

So if we think about the first one, we need to prove that $||\gamma(\phi(x))||=||\phi(x)||$(Am I correct?) 
So, $||\gamma(\phi(x))||=\sup_{||x+L||=1}|\phi(x+L)|$ and $||\phi(x)||=\sup_{||x||=1}|\phi(x)|$. Moreover, $||x+L||=\inf_{y\in L}||x+y||$. But now I don't see a way to compare $||\gamma(\phi(x))||$ and $||\phi(x)||$. Is my approach correct? Can somebody please help me to proceed?

*

*We have that
$$\Vert \gamma(\phi) \Vert =\sup_{\Vert x+L\Vert=1}\vert\phi(x+L)\vert \leq \sup_{\Vert x+L\Vert=1}\Vert\phi \Vert \Vert x+L\Vert = \Vert\phi\Vert.$$
Now you just need to show that "$\geq$" holds.

*Due to the Hahn-Banach-theorem $\beta$ is surjective. For all $\phi \in X^*$ we have
$$\Vert \phi  + L^a \Vert= \inf_{\psi \in L^a} \Vert \phi  + \psi \Vert \geq \inf_{\psi \in L^a} \Vert (\phi  + \psi)|_L \Vert = \Vert \phi |_L \Vert = \Vert \beta(\phi  + L^a )\Vert. $$
Using Hahn-Banach-theorem there exist an extension $\Phi \in X^*$ of $\phi |_L$ with $\Vert \Phi \Vert = \Vert \phi|_L \Vert$. Hence 
$$\Vert \phi  + L^a  \Vert \leq \Vert \phi  + \underbrace{(\Phi - \phi )}_{\in L^a}\Vert = \Vert \Phi \Vert = \Vert \phi |_L \Vert = \Vert \beta(\phi  + L^a )\Vert.$$
Hence $\beta$ is isometric. 
I hope that it helps you :)
