Let $A$ be a normed space and $B\subset A$ a linear subspace with $\phi\in B^*$.
I am looking to prove the following:
B is dense in $A$ iff there is a unique extension of $\phi$ to a continuous linear functional on $A$.
What I thought:
Hahn-Banach tells us that there does exist an extension of $\psi$ of $\phi$ such that $\|\phi\|=\|\psi\|$.
Further $B$ being dense in $A$ means that each sequence in $B$ converges to an $a\in A$. How do I use these to prove the statement?
Edit:
$\implies$: Suppose $f,g$ are extensions of $\phi$ on $A$. Let $a\in A\backslash B$. Then $a=\lim_{n\rightarrow\infty}b_n$ for $b_n\in B$. This gives
$$f(a)=f(\lim_{n\rightarrow\infty}b_n)=\lim_{n\rightarrow\infty}f(b_n)=\lim_{n\rightarrow\infty}g(b_n)=g(\lim_{n\rightarrow\infty}b_n)=g(a).$$
Is this a good proof? How does the other way go?
Best Answer
Suppose $B$ is dense and $f$ and $g$ are extensions of $\phi$, $f-g$ vanishes on $B$ so it vanishes on its adherence, thus $f=g$ and the extension is unique.
On the other hand,suppose there exists a unique extension. If $B$ is not dense, consider the adherence $L$ of $B,$ there exists an element $x$ not in $L$. We denote by $M$ the vector subspace generated by $x$ and $L$. Let $f$ be the linear form defined on $M$ whose restriction to $L$ is zero and $f(x)=1$. Suppose that $f$ is not bounded. There exists a sequence $y_n,$ of norm 1 such that $,lim_nf(y_n)=+\infty$. Write $,y_n=t_nx+l_n$, we have $f(y_n)=t_n$. We deduce that $lim_nt_n=+\infty$. This implies that $lim_nl_n/t_n=-x$. Contradiction since ,$L$ is closed and $x$ not in $L$. Hahn Banach implies that we can extend $f$ to the linear function $g$ defined on $A$. Remark that the restriction of of $h$ to $B$ is zero. We deduce that the zero function on $B$ has two different extensions. Contradiction.