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Let $X$ be a normed vector space and let $Y$ a subspace of $X$. Prove that $X^*/Y^0$ is isometrically isomorphic to $Y^*$, where $$Y^0=\{y^* \in Y^*|y^*(y)=0, \forall y \in Y \}.$$

Proof. Define $T\colon X^*/Y^0 \longrightarrow Y^*$ by $T(x^*+Y^0)=x^*$. Then $T$ is well-defined. Indeed, suppose that $$x_1^*+Y^0=x_2^*+Y^0.$$ Then $x_1^*-x_2^* \in Y^0.$ Consequently, $$x_1^*(y)=x_2^*(y), \forall y \in Y \Longrightarrow T(x_1^*+Y^0)=T(x_2^*+Y^0).$$

Also $T$ is an isometry. Indeed $$\begin{array}{lcl}\|T(x^*+Y^0)\|&=&\|x^*\|\\ &=&\sup\{|x^*(y)|:\|y\| \leqslant 1\}\\ &=&\sup\{|x^*(-y)|:\|-y\| \leqslant 1, y \in Y\}\\ &=&\sup\{|-x^*(y)|:\|y\| \leqslant 1,y \in Y\}\\ &=&\inf\{\|-x^*-z^*\|:z^* \in Y^0\}\\ &=&\inf\{\|x^*+z^*\|:z^* \in Y^0\}\\ &=&\|x^*+Y^0\|_{X^*/Y^0}.\end{array}$$

To prove that T is an isometry, I used one of the consequences of the Hahn-Banach theorem which states:

Let $X$ a normed linear space and $Y$ a subspace of $X$ and $x^* \in X^*$.Then $$\sup\{|x^*(y)|:\|y\| \leqslant 1,y \in Y \}=\inf\{\|x^*-z^*\|:z^* \in Y^0\}.$$

Finally its obvious that $T$ is onto.

Is this proof correct or am i missing something?

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