Let $V$ a closed subspace of a normed space $X$. Prove that the following aplication is a isometric isomorphism.

Question

Let $\pi_V:X\rightarrow\dfrac{X}{V}$, where $\pi_V(x)=x+V$ and $$p:\left(\dfrac{X}{V}\right)'\rightarrow V^0,\quad p(f)=f\pi_V.$$ Prove that $p$ is a isometric isomorphism. Where, $V^0=\{f\in X':f|_S\equiv0 \}$. I proved that $p$ is a isomorphism, and the continuity of $p$, we had $$||p(f)||=||f\circ \pi_V||\leq ||f||\cdot||\pi_V||=||f||,$$ because, if $V\neq X$ we had $||\pi_V||=1$. Now, remains to show that $||p(f)||\geq||f||$.

Answer 1

To make sense out of this you have to identify $(\frac X V)' $ with $V^{0}$. A continuous linear functinal $f$ on $\frac X V$ is identified with the element $F$ of $X'$ which takes $x$ to $f(x+V)$ or $(f\circ \pi_V) (x)$. Thus $F=p\circ f$. Let $\epsilon>0$. There exist $x$ such that $\|x+V\| \leq 1$ and $|f(x+V)| >\|f\|-\epsilon$. Now $\|p(f)\|=\|F\| \geq |F(x)|=|f(x+V)| >\|f\|-\epsilon$ and $\epsilon$ is arbitrary.