Uniqueness of Hahn-Banach extension for dense subspace

Question

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?

Answer 1

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.

Answer 2

Let $B$ be a dense linear subspace of $A$ and $\phi:B\to\mathbb K$ a continuous functional. This means $\|\phi\|_B$ exists. First we show that denseness implies the existence of an extension.

Since we are a in a normed vector space situation the closure of $B$ is the same as the sequential closure of $B$, meaning any point in $A$ can be approximated by a sequence $b_n$ in $B$. Now if $b_n\to a\in A$ you have that $$\|\phi(b_n)-\phi(b_m)\|≤\|\phi\|\cdot\|b_n-b_m\|$$ and the sequence $\phi(b_n)$ is a Cauchy-sequence. Since $\mathbb K$ is complete there exists a limit, define $\phi(b)$ to be this limit. It is necessary to show that this definition is well defined, ie if $b_n\to b$ and $\tilde b_n\to b$ that then $$\lim_n\phi(b_n)=\lim_n\phi(\tilde b_n)$$ This follows because $b_n-\tilde b_n\to0$ and for that reason $\phi(b_n)-\phi(\tilde b_n)=\phi(b_n-\tilde b_n)\to0$ from continuity. So any functional $\phi: B\to\mathbb K$ has a continuous extension onto $A$. This does not use Hahn Banach.

If two continuous $f_1,f_2: X\to Y$ functions (where $Y$ is a metric space) agree on a dense subset of $X$ then they must be equal. This shows that the extension has to be unique.

($\Delta(x)=d(f_1(x),f_2(x))$ is a continuous function, then $\Delta^{-1}(\{0\})=\{x\mid f_1(x)=f_2(x)\}$ is closed.)

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For the other direction, suppose $B$ is not dense in $A$, ie there exists a point $x\in A$ so that $$d(x,\overline{B})=\inf_{y\in B}d(x,y)>0.$$ Then there exist different extensions of a linear functional $\phi:B\to\mathbb K$ onto $B\oplus\text{span}\{x\}$, for example: $$\phi_1(b+\lambda x)=\phi(b)\qquad \phi_2(b+\lambda x)=\phi(b)+\lambda$$

In order to check well definedness, note that the norm of $b+\lambda x$ is larger than $\|b\|+|\lambda|\,d(\overline B,x)$. This is because $$\|b+\lambda x\|=d(b,-\lambda x)≥d(\overline{B},-\lambda x) =|\lambda| d(\overline{B},x)$$ For that reason $\|\phi_1\|≤\|\phi\|$, whereas $$\|\phi_2(b+\lambda x)\|=|\phi(b)+\lambda|≤\|\phi\|\cdot\|b\|+|\lambda|≤\frac{\|\phi\|}{d(\overline B,x)} \left(d(\overline B,x)\|b\|+ \frac{d(\overline B,x)}{\|\phi\|}\, |\lambda| \right)$$ The term on the right can be made even bigger by replacing $$d(\overline B,x)\|b\|\to A\|b\|\\ \frac1{\|\phi\|}\,|\lambda|\to A|\lambda|$$ Where $A:=\max(d(\overline B,x),\frac1{\|\phi\|})$. Put it in to find $$\|\phi_2(b+\lambda x)\|≤\frac{A\|\phi\|}{d(\overline B,x)}\left(\|b\|+d(B,\overline x)|\lambda|\right)≤\frac{A\|\phi\|}{d(\overline B,x)}\|b+\lambda x\|$$

Now apply Hahn-Banach.

Answer 3

For the "other direction" it suffices to show that if $B$ is not dense in $A$ then $\phi$ has 2 unequal extensions to members of $A^*.$ By the Hahn-Banach Theorem let $\phi_1\in A^*$ be an extension of $\phi.$ The idea is to find $g\in A^*$ with $g(\bar B)=\{0\}$ and $g\ne 0,$ and let $\phi_2=\phi_1+g.$

Take $x\in A$ \ $\bar B.$ Let $C$ be the vector subspace generated by $\{x\}\cup \bar B.$ Every $y\in C$ is equal to $sx-y$ for a unique scalar $s,$ and a unique $y\in \bar B.$ (Because if $y,y'\in \bar B,$ and $sx-y =s'x-y'$ then $s'\ne s \implies x=(y-y')/(s-s')\in \bar B.)$

Let $d=\inf\{\|x-z\|:z\in \bar B\}.$ Then $d>0$ because $x\not \in \bar B.$

For $sx-y\in C$ let $f(sx-y)=sd.$

Now for $y\in \bar B$, if $s\ne 0$ we have $|f(sx-y)|=|s|\cdot d\leq |s|\cdot \|x-(y/s)\|=\|sx-y\|$, while if $s=0$ we have $f(sx-y)=0.$ So $f\in C^*.$

Apply Hahn-Banach to extend $f$ to $ g\in A^*.$ Note that $g(x)=f(x)=d \ne 0$ and that $g(y)=f(y)=0$ for all $y\in \bar B.$