Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 7 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.