So I'm pondering the statement:

Show that a continuous linear functional $f$ on a subspace $V$ of a Hilbert space $H$ has a unique norm preserving extension $h$ on $H$

Here is my thought.

Consider the closure of $V$, $\overline{V}$. Then since $V$ is dense in $\overline{V}$, there exists a unique extension $\overline{f}$ of $f$ to $\overline{V}$.

Moreover since $\overline{V}$ is a closed subspace of $H$ then it is also a Hilbert space, so by the Frechet-Riesz theorem we have $\overline{f}(v)$ = $\langle v,y_{\space\overline{f}} \rangle$ for some unique element $y_{\space\overline{f}} \in \overline{V}$

Take $ h(x) = \langle x,y_{\space\overline{f}} \rangle$.

How do I get uniqueness though?


1 Answer 1


Let me simplify things by assuming (as you point out, without loss of generality) that $V$ is closed. Then, indeed, $f(x) = \langle x,y \rangle$ on $V$ for a unique $y \in V$, yielding the immediate extension $h(x) := \langle x,y\rangle$ of $f$ to $H$; one can readily check that $$\|f\|_{V^\ast} = \|y\|_V = \|y\|_H = \|h\|_{H^\ast},$$ so that this extension is norm-preserving.

Now, suppose that $k \in H^\ast$ is another extension of $f$ to $H$, so that $k(x) = \langle x,z\rangle$ for a unique $z \in H$. Then for any $x \in V$, $$0 = f(x) - f(x) = h(x) - k(x) = \langle x,y\rangle - \langle x,z\rangle = \langle x,y-z\rangle,$$ so that $y-z \in V^\perp$. Hence, $y$ and $y-z$ are orthogonal, so that $$\|k\|_{H^\ast} = \|z\|_H = \|y - (y-z)\|_H = \sqrt{\|y\|^2_H + \|y-z\|^2_H}.$$ Thus, $\|k\|_{H^\ast} = \|f\|_{V^\ast}$ if and only if $\sqrt{\|y\|^2_H + \|y-z\|^2_H} = \|y\|_H$, if and only if $\|y-z\| = 0$, if and only if $y =z$, if and only if $h=k$. Hence, $h$ is indeed the unique norm-preserving extension.

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    $\begingroup$ Actually, there seems to be a small typo in the second sentence: it says "yielding the immediate extension .. of $f$ to $V$", whereas it should be "yielding the immediate extension .. of $f$ to $H$". $\endgroup$
    – Olorun
    Commented Sep 17, 2015 at 4:21
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    $\begingroup$ On the one hand, by definition, $y \in V$. On the other hand, by the second displayed equation, $y-z \in V^\perp$. $\endgroup$ Commented Jun 25, 2018 at 13:44
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    $\begingroup$ The vector $y \in V$ is still a vector in the ambient Hilbert space $H$, so there’s nothing to stop you from taking the inner product of any vector in $H$ with $y$, and this is the extension $h$. $\endgroup$ Commented Jan 27, 2019 at 14:15
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    $\begingroup$ Note that if $w \in V^\perp$, then $k(x) := \langle x, y+w\rangle$ defines another extension of $f$ to a functional on $H$. The whole point is that $h$ is the unique extension of $f$ with the same operator norm as $f$; if $w \neq 0$, then $k$ will actually have a larger operator norm that $h$. All this is discussed in my answer—please re-read it carefully. $\endgroup$ Commented Jan 27, 2019 at 16:39
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    $\begingroup$ That's correct. $\endgroup$ Commented Jan 27, 2019 at 18:15

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