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First of all, here is the assignment:

Let $X$ be a Hilbert space over $\mathbb{C}$, $V \subseteq X$ be a closed subspace and $f \in L(V, \mathbb{C}) $ a linear continuous operator. Show that there exists one unique continuation $F$ of $f$ on $X$, such that all these properties are satisfied:

  1. $F \in L(X, \mathbb{C})$ (i.e., linear and continuous),
  2. $F|_V = f$,
  3. $\|F\| = \|f\|$.

There is also a hint given: Use the Riesz representation theorem. Ok, so I should know where to start, but fact is: I have no real clue. I know that using Riesz theorem property 3 can easily be shown from property 2. But how do I prove that there exists this continuation? I'd be glad if anyone could give me a hint on how to approach this.

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Since $V$ is closed, $V$ endowed with the inner product of $X$ is a Hilbert space. So $f$ and be represented as $f(\cdot)=\langle v_0,\cdot\rangle$. Now use this formula to extend it to $X$ and check that all properties are satisfied. –  Davide Giraudo Dec 20 '11 at 10:32
    
It is also true if $V$ is not closed. –  Jonas Meyer Dec 21 '11 at 6:30
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1 Answer

up vote 6 down vote accepted

Since $V$ is a closed subspace of $X$, it is a Hilbert space too. By the Riesz representation theorem $f$ is of the form $f(v) = \langle v, \xi \rangle$ for a unique $\xi \in V$. Now note that $F(x) = \langle x, \xi\rangle$ is defined for all $x \in X$ and satisfies all the requirements. I'll leave uniqueness of the extension $F$ to you.

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