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A similar question has been posted here: Dense subspace of linear space and functional equal to $0$.

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 $$^{\bot}N:=\{x\in V:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ f\in N\}=\bigcap_{f\in N}\ker(f)\subset V.$$

I am working on a problem stating as follows:

Let $W\subset V$ be a vector subspace. Show that $W$ is dense in $V$ if and only if $W^{\bot}=\{0\}$.

Using Hahn-Banach geometric form, I have proved that $\overline{W}=\ ^{\bot}(W^{\bot})$. This gives us the direction $(\Leftarrow)$, as follows:

Suppose $W^{\bot}=\{0\}$, then $$\overline{W}=\ ^{\bot}(W^{\bot})=\ ^{\bot}\{0\}=\{x\in V: \langle x,0\rangle_{V,V^{*}}=0\}=V,$$ since $\langle x,0\rangle_{V,V^{*}}=0$ is always true for all $x\in V$. So, we have $\overline{W}=V$, which means that $W$ is dense in $V$.

However, I got stuck in the direction $(\Rightarrow)$. Suppose $W$ is dense in $V$, and suppose $W^{\bot}\supset \{0\}$, i.e. there exists $f\in V^{*}$ that is not identically zero such that $\langle x, f\rangle_{V,V^{*}}=0$ for all $x\in W$.

I tried to use the density to continue, but by density we can only $y\in V$ such that $\|x-y\|_{V}<\epsilon$, where $\epsilon>0$ was arbitrarily fixed. I don't know how to continue then...

Is there a way for me to get contradiction? Thanks!

JacobsonRadical
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1 Answers1

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Assume there is an $f\ne 0$ which vanishes on $W$. Then at some point $v\in V$ we have $f(v)\ne 0$. Since $W$ is dense, we find a sequence $(w_n)$ inside $W$ converging to $v$. By continuity, $$ f(v)=f(\lim w_n)=\lim f(w_n)=\lim 0=0\ . $$A contradiction.

$\square$

dan_fulea
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  • If using the language of inner product, the sequence $(w_{n})$ converging to $v$ in the sense that $|w_{n}-v|{V}\longrightarrow 0$ as $n\rightarrow\infty$. Then, how could I translate this to $\langle v, f\rangle{V,V^{*}}$? – JacobsonRadical Mar 03 '21 at 15:30
  • I suppose $\langle v, f\rangle_{V,V^*}$ is the "simpler way" to write $f(v)$. If not, please define the pairing. – dan_fulea Mar 03 '21 at 15:32
  • No worry. I over-thought. Thanks for the great answer! – JacobsonRadical Mar 03 '21 at 15:33
  • At any rate, i would expect something like $|\langle v,f\rangle|\le |f|;|v|$. – dan_fulea Mar 03 '21 at 15:34
  • yes, I used something like this: we have $|w_{n}-v|{V}\longrightarrow 0$ as $n\rightarrow\infty$, and we use $$|\langle w{n}, f\rangle_{V,V^{}}-\langle v, f\rangle_{V,V^{}}|\leq |w_{n}-v|\cdot|f|{op},$$ to conclude $\langle w{n}, f\rangle\longrightarrow \langle v,f \rangle,$ and things follow. – JacobsonRadical Mar 03 '21 at 15:36