How does the Hahn-Banach theorem for Hilbert spaces follow from Riesz's representation theorem?


1 Answer 1


Yes. Given a subspace $M$ of a Hilbert space $H$, and a continuous linear functional $f : M \to \mathbb{C}$, you can extend $f$ to a unique continuous linear functional $g : \overline{M} \to \mathbb{C}$.

Now, $\overline{M}$ is a Hilbert space, to there is $y \in \overline{M}$ such that $$ g(x) = \langle x, y\rangle $$ for all $x\in \overline{M}$.

Now write $h : H \to \mathbb{C}$ to be $x \mapsto \langle x, y\rangle$ and this extends $f$.

  • $\begingroup$ How to show the uniqueness?? @Prahlad $\endgroup$
    – user487904
    Nov 22, 2017 at 9:33
  • $\begingroup$ There's no uniqueness in the Hahn-Banach theorem. Also, if $H$ is finite-dimensional, there can be more than one continuous linear functional extending $f$, and the norm of a continuous linear extension can be larger than $f$. To see this, extend a basis of $M$ to a basis of $H$, then extend $f$ to the basis of $H$ arbitrarily. $\endgroup$ Sep 27, 2018 at 16:50
  • $\begingroup$ @Prahlad Is a closed subspace of a Hilbert space always a Hilbert space? $\endgroup$
    – user974406
    Dec 1, 2021 at 19:34
  • $\begingroup$ @user974406 Yes, this is a general fact. A subspace of a complete metric space is complete (e.g. a subspace of a Hilbert space is Hilbert) iff it is closed under the standard topology. $\endgroup$ Apr 7 at 7:04

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