I am reading Haim Brezis's functional analysis, and I am confused by on of his proof.
Let $(V, \|\ \cdot\ \|)$ be a Banach space on $\mathbb{C}$ and let $V^{*}$ be its dual, endowed with the dual norm $$\|f\|_{V^{*}}=\|f\|_{op}=\sup_{\|x\|_{V}\leq 1}|\langle x,f \rangle_{V, V^{*}}|.$$ For $M\subset V$, we define the annihilator of $M$ as $$M^{\bot}:=\{f\in V^{*}:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ x\in M\}\subset V^{*}.$$ and similarly for $N\subset V^{*}$, we define $$^{\bot}N:=\{x\in V:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ f\in N\}=\bigcap_{f\in N}\ker(f)\subset V.$$
Proposition 1.9 states the following:
Let $M\subset V$ be a linear subspace. Then, $$^{\bot}(M^{\bot})=\overline{span\ M}.$$
I have proved $\overline{span\ M}\subset ^{\bot}(M^{\bot})$, and the other direction $\overline{span\ M}\supset ^{\bot}(M^{\bot}$ is an application of Hahn-Banach theorem geometric form. The proof goes as follows:
Suppose by contradiction that there exists some $x_{0}\in ^{\bot}(M^{\bot}$ such that $x_{0}\notin \overline{span\ M}.$ It is clear that $\{x_{0}\}$ and $\overline{span\ M}$ then has empty intersection, both convex, and the first one is compact, the second one is closed.
Then, by Hahn-Banach theorem, second geometric form, there is a closed hyperplane that strictly separates $B=\{x_{0}\}$ and $A=\overline{span\ M}$. That is, there exists some $f\in V^{*}$ and $\alpha\in\mathbb{R}$ such that $$\langle x,f\rangle_{V,V^{*}}<\alpha<\langle x_{0}, f\rangle_{V,V^{*}}\ \ \text{for all}\ \ x\in \overline{span\ M}.$$
So far so good. But then, Brezis claimed that
Since $M$ is a linear space, it follows that $\langle x, f\rangle_{V,V^{*}}=0$ for all $x\in M$. Why this is true? I thought about this for a while but don't follow why linear space implies this result…
The following is the statement of Hahn-Banach 2nd geometric form:
Let $A,B\subset V$ be two nonempty convex subsets such that $A\cap B=\varnothing$. Assume that $A$ is closed and $B$ is compact. Then, there exists a closed hyperplane that strictly separates $A$ and $B$. That is, there exists some $f\in V^{*}$ and $\alpha\in\mathbb{R}$ such that $$\langle x,f\rangle<\alpha\ \ \text{for all}\ \ x\in A\ \ \text{and}\ \ \langle x,f\rangle>\alpha\ \ \text{for all}\ \ x\in B.$$
Best Answer
Replace $x$ by $\beta x$ in your inequality, with arbitrary $\beta$.
More abstractly, the image of $f$ is a subspace of $\mathbb{R}$ because $f$ is linear, and you already know it's not all of $\mathbb{R}$ from your inequality.