I am struck on a problem in my operator theory class so I am posting it here in the hope of getting some help.
Let $H$ be a Hilbert space and let $E\subseteq H$ be a closed subspace. Show that if $\phi \in E^{*}$ , then $\phi$ has a unique Hahn Banach extension $\psi \in H^{*}$.
As $H$ is a hilbert space and so a topological vector space which is locally convex so the Hahn-Banach theorem is valid. The theorem as given in my lectures is : Let $X$ is a topological vector space which is locally convex over $\mathbb{K} \in $ {$\mathbb{R} , \mathbb{C}$} . Let $E$ be a subspace of $X$ and $\phi^{'}: E\to \mathbb{K}$ be a subspace of $X$ and $\phi: E\to \mathbb{K}$ continuous linear form. Then there exists a continuous linear form $\phi|_{E} =\phi^{'} $.
Using the above theorem , it can be deduced that the extension exists.
Now let there exists 2 extension $\phi$ and $\psi$. Now, as a Hilbert space is a normed space. So, such extensions will satisfy the following ( by a corollary of Hahn-Banach theorem): $||\phi ||=||\phi^{'} || =||\psi||$.
But I don't have any ideas on how to prove the equality.
Kindly help!
Best Answer
For any $x\in H$ there are unique $x_1,x_2$ such that $x=x_1+x_2$ with $x_1\in E$ and $x_2\in E^{\perp}$ (i.e. the inner product $(x_2,z)=0$ for every $z\in E).$ We have $x\in E\iff x=x_1.$ And $\|x\|^2=\|x_1\|^2+\|x_2\|^2$ for all $x.$
The case $\phi=0$ is trivial.
For $\phi\ne 0,$ suppose $0<r=\sup \{\frac {|\psi (y)|}{\|y\|}:0\ne y\in E^{\perp }\}.$ Then $$(1).....\|\psi\|\ge \sup \left\{\frac {|\phi (x)+s\cdot \psi (y)|}{\|x+sy\|}: x\in E\land y\in E^{\perp }\land \|x\|=\|y\|=1\land s>0\right\}\ge$$ $$(2)....\ge\sup \left\{\frac {|\phi (x)+s\cdot \psi (y)|}{1+s^2/2}: x\in E\land y\in E^{\perp }\land \|x\|=\|y\|=1\land s>0\right\}\ge$$ $$(3)....\ge\sup \left\{\frac{|\phi (x)|+s\cdot |\psi (y)|}{1+s^2/2}: x\in E\land y\in E^{\perp}\land \|x\|=\|y\|=1\land s>0\right\}=$$ $$(4).... =\|\phi\|+r\cdot\sup_{s>0}\frac {s}{1+s^2/2}>\|\phi\|,$$ contrary to $\|\psi\|=\|\phi\|.$
Therefore $\psi(y)=0$ for all $y\in E^{\perp}.$ So for all $x,$ with $x_1,x_2$ as in the 1st sentence above, we have $\psi(x)=\psi(x_1)+\psi (x_2)=\psi(x_1)=\phi(x_1).$
Explanation:(i). The denominator in $(1)$ satisfies $\|x+sy\|=(1+s^2)^{1/2}<1+s^2/2$....(ii). The transition from $(2)$ to $(3)$ is justified because in $(3)$ we are restricting the set of $x,y$ to a smaller set, namely those for which $|\phi(x)+s\cdot\psi(y)|=|\phi(x)|+s\cdot|\psi(y)|.$ Also, to get from $(3$ to $(4)$, for every $x\in E, y\in E^{\perp}$ there are scalars $k, k'$ such that $|k|=|k'|=1$ and $|\phi(kx)+s\cdot\psi(k'y)|=|\phi(x)|+s\cdot|\psi(y)|.$