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Let $(\mathcal{H},\langle\cdot,\cdot\rangle)$ be a Hilbert Space, $U\subset \mathcal{H},U\not=\mathcal{H}$ be a closed subspace and $x\in\mathcal{H}\setminus U$. Prove that there exists $\phi\in\mathcal{H}^*$, such that\begin{align}\text{Re } \phi(x)<\inf_{u\in U}\text{Re }\phi(u) \end{align} Hint: Observe that $\inf_{u\in U}\text{Re }\phi(u)\leq0$.

This seems like an application of the Banach Seperation theorem. But the way I know it is not directly applicable. I know that for two disjoint convex sets $A$ and $B$ of which one is open there exists a functional seperating them. Is there anything special in this problem about $\mathcal{H}$ being Hilbert and not some general Banach space?

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ok. This yields $\text{Re }\phi(x)=\phi(x)=-\lVert x\lVert$. Also $\text{Re } \phi(u)=-\text{Re } \langle u,x\rangle\geq-\lvert\langle u,x\rangle\lvert\geq -\lVert u\lVert\lVert x\lVert$. For $\lVert u\lVert<1$ this implies $\text{Re }\phi(u)>\text{Re }\phi(x)$ but what about $u\geq1$? –  Julian Jul 14 '12 at 21:40
    
Ah, my mistake, Norbert has corrected this. (Although for what it's worth $\langle x,x\rangle = \|x\|^2$, and that $|\langle u,x\rangle|$ is strictly less than $\|u\|\|x\|$ for all $u\in U$.) –  Jonas Meyer Jul 14 '12 at 21:43

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up vote 2 down vote accepted

There is more general result which proof you can find in Rudin's Functional Analysis

Let $A$ and $B$ be disjoint convex subsets of topological vector space $X$. If $A$ is compact and $B$ is closed then there exist $\varphi\in X^*$ such that $$ \sup\limits_{x\in A}\mathrm{Re}(\varphi(x))<\inf\limits_{x\in B}\mathrm{Re}(\varphi(x)) $$

Your result follows if we take $X=\mathcal{H}$, $A=\{x\}$ and $B=U$.

Of course this is a sledgehammer for such a simple problem, because in case of Hilbert space we can explicitly say that functional $$ \varphi(z)=\langle z, \mathrm{Pr}_U(x)-x\rangle $$ will fit, where $\mathrm{Pr}_U(x)$ is the unique orthogonal projection of vector $x$ on closed subspace $U$.

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