Let $X$ be a normed space and $Y$ a closed subspace of $X$ and $Y^0=\{f \in X^*|f(x)=0,\forall y \in Y\}$.
Prove that $Y^0$ is isometricaly isomorphic with $(X/Y)^*$.
I have to find a function specifically between $Y^0 $ and $(X/Y)^*$.
One idea is the function $S: Y^0 \longrightarrow (X/Y)^*$ such that $S_{f}(x+Y)=f(x)$ but i have a difficult time to proved that it is bounded and an isometry.
Can someone help me please or give another idea for an bijection?
Thnak you in advance.
We assume that $Y \neq X$.
Let $x+ Y \in X/Y$ with $\lVert x+Y \rVert \leq 1$. Let $\epsilon > 0$. Then, there is $y \in Y$ with $\lVert x-y \rVert \leq 1 + \epsilon$. Now $$ \lvert S(f)(x+Y)\rvert = \lvert f(x) \rvert = \lvert f(x-y)\rvert \leq (1+ \epsilon) \lVert f \rVert. $$ This shows that $\lVert S \rVert \leq 1$. To show that $\lVert S \rVert \geq 1$ pick some $x \in X \setminus Y$ with $\lVert x \rVert \leq 1$ (this ensures $\lVert x + Y \rVert \leq 1$). Now we apply a standard Hahn-Banach argument: There exists a functional $f \in X^*$ with $f(Y) = \{0\}$, $f(x) = 1$ and $\lVert f \rVert \leq 1$. Then $\lvert S(f)(x +Y) \rvert = \lvert f(x) \rvert = 1$ and hence $\lVert S(f) \rVert \geq 1$ and so $\lVert S \rVert \geq 1$.
It is clear, that $S$ is bijective.