Theorem: Annihilator of a $Y$ a subspace of $X$

Question

If $X$ is a normed vector space and $Y$ is a closed subspace, then $X^{*}/Y^{0}$ is isometrically linearly isomorphic to $Y^{*}$.

Here, $X^{*}$ denotes dual space of $X$; $Y^{0}$ denotes annihilator of $Y$.

Please provide some hints to prove this statement. I have been able to prove that for an $f\in Y^{*}$ there exists an $f_{\text{ext}}\in X^{*}$ with the same norm, using Hahn-Banach theorem. But I am struggling with proving the linearity part. For example, if $f,g\in Y^{*}$ with norms $\|f\|$ and $\|g\|$, and have extensions $f_{\text{ext}}$ and $g_{\text{ext}}$. Does $f+g$ has the extension $f_{\text{ext}}+g_{\text{ext}}$ with same norm as $\|f+g\|$.

Note: The same question has been asked at annihilator subspace of normed space. But it does not contain a reasonable answer.

Answer 1

Define $T(x^{\ast}+Y^{\circ})(y)=x^{\ast}(y)$, we are to show that $T$ is isometric isomorphism.

For well-definedness: $x^{\ast}+Y^{\circ}=z^{\ast}+Y^{\circ}$, then $x^{\ast}-z^{\ast}\in Y^{\circ}$ and so $(x^{\ast}-z^{\ast})(y)=0$ and hence $x^{\ast}(y)=z^{\ast}(y)$.

For surjectivity: For $y^{\ast}\in Y^{\ast}$, Hahn-Banach Extension Theorem says that some $x^{\ast}$ is such that $x^{\ast}\big|_{Y}=y^{\ast}$, then $T(x^{\ast}+Y^{\circ})(y)=x^{\ast}(y)=y^{\ast}(y)$.

For isometric: Let $y^{\ast}=T(x^{\ast}+Y^{\circ})$, let $x_{y^{\ast}}^{\ast}$ be one of its Hahn-Banach extension, then $x_{y^{\ast}}^{\ast}+Y^{\circ}=x^{\ast}+Y^{\circ}$, if $z^{\ast}\in Y^{\circ}$, then $y^{\ast}=x^{\ast}+z^{\ast}$ on $Y$, so \begin{align*} \|y^{\ast}\|&=\sup\{|(x^{\ast}+z^{\ast})(y)|: y\in B_{Y}\}\\ &\leq\sup\{|(x^{\ast}+z^{\ast})(x)|: x\in B_{X}\}\\ &=\|x^{\ast}+z^{\ast}\|, \end{align*} so $\|y^{\ast}\|\leq\inf\{\|x^{\ast}+z^{\ast}\|: z^{\ast}\in Y^{\circ}\}=\|x^{\ast}+Y^{\circ}\|=\|x_{y^{\ast}}^{\ast}+Y^{\circ}\|\leq\|x_{y^{\ast}}^{\ast}\|\leq\|y^{\ast}\|$, so $\|x^{\ast}+Y^{\circ}\|=\|y^{\ast}\|=\|T(x^{\ast}+Y^{\circ})\|$.

Answer 2

Let $(f+g)_{\text{ext}}$ be a Hahn-Banach extension of $f+g$.

Notice that $(f+g)_{\text{ext}}$ and $f_{\text{ext}} + g_{\text{ext}}$ are both equal to $f+g$ when restricted to $Y$.

Therefore

$$(f+g)_{\text{ext}} - (f_{\text{ext}} + g_{\text{ext}}) \in Y^0$$

so we can conclude $$(f+g)_{\text{ext}} + Y^0 = f_{\text{ext}} + g_{\text{ext}} + Y^0$$

Similarly $(\lambda f)_{\text{ext}}$ and $\lambda f_{\text{ext}}$ are both equal to $\lambda f$ on $Y$ so $$(\lambda f)_{\text{ext}} + Y^0 = \lambda f_{\text{ext}} + Y^0$$

Finally, by definition:

$$\|f_{\text{ext}} + Y^0\|_{X^*/Y^0} = \inf_{h \in Y^0} \|f_{\text{ext}} - h\|_{X^*} $$

$$\inf_{h \in Y^0} \|f_{\text{ext}} - h\|_{X^*} \le \|f_{\text{ext}}\|_{X^*} = \|f\|_{Y^*}$$

For any $h \in Y^0$ we have: $$\|f_{\text{ext}} - h\|_{X^*} \ge \left\|(f_{\text{ext}} - h)|_Y\right\|_{Y^*} = \|f\|_{Y^*}$$

so we conclude $\|f_{\text{ext}} + Y^0\|_{X^*/Y^0} = \|f\|_{Y^*}$.

Therefore, the map $Y^* \to X^*/Y^0$ given by $$f \mapsto f_{\text{ext}} + Y^0$$ where $f_{\text{ext}}$ is a Hahn-Banach extension of $f$ to $X$, is a well-defined linear isometry.

For surjectivity, consider $F + Y^0$ where $F \in X^*$.

We have that $F$ and $(F|_Y)_{\text{ext}}$ are both equal on $Y$ so

$$(F|_Y)_{\text{ext}} + Y^0 = F + Y^0$$

hence $$F|_Y \mapsto F + Y^0$$