Here is the proof that every Hilbert space is refexive:

Let $\varphi\in\mathcal{H^{**}}$ be arbitrary. By Riesz, there is a unique $f_\varphi\in\mathcal{H^*}$ with

$\varphi(f)=\langle\,f,f_\varphi\rangle$ for all $f \in\mathcal{H^*} $.

Using the same notation and theorem, we have

$\hat{y}_{f_\varphi}(f)= f(y_{f_\varphi})=\langle\,y_{f_\varphi},y_f\rangle=\langle\,f,f_\varphi\rangle=\varphi(f)$

This implies $\hat{y}_{f_\varphi}=\varphi$, thus $\mathcal{H}$ reflexive.

I understood all the steps except for the last implication. Basically, we just showed that $2$ functionals from bi-dual space $\mathcal{H^{**}}$ are the same, why would it imply that $\mathcal{H}$ is reflexive? Any explanation would be highly appreciated!

  • $\begingroup$ Remember that $\mathcal H$ is reflexive if the canonical injective map from $\mathcal H$ to $\mathcal H^{**}$ is also surjective $\endgroup$ Commented Aug 16, 2019 at 16:08
  • $\begingroup$ @Omnomnomnom I'm awared of that, but can't still get the idea somehow .. Could you please give more precise explanation? This is not an exercise by the way, just preparation for the exam :) $\endgroup$
    – Newbie001
    Commented Aug 16, 2019 at 16:10
  • $\begingroup$ There's a missing item in the last equation. The leftmost term should be $\hat y_{f_\varphi} {\bf(f)} =f(y_{f_\varphi})=\dots$. $\endgroup$
    – Berci
    Commented Aug 16, 2019 at 16:31
  • $\begingroup$ @Berci yes, you're right! I've just edited. Thanks! $\endgroup$
    – Newbie001
    Commented Aug 16, 2019 at 16:56

2 Answers 2


Let $\Phi:\mathcal H \to \mathcal H^{**}$ denote the canonical injection, AKA the evaluation map (in the notation of the proof, $\Phi(x) = \hat x$)t. We want to prove that $\Phi$ is surjective. In other words: we want to prove that for any $\varphi \in \mathcal H^{**}$, there exists a $y \in \mathcal H$ such that $\Phi(y) = \varphi$.

So, begin with any $\varphi$. By the RRT, there exists a unique $f_{\varphi}$ such that for all $f \in \mathcal H^*$, $\varphi(f) = \langle f, f_{\varphi}\rangle$.

Note that this requires that requires an inner product on $\mathcal H^*$. Recall how such an inner product is defined: RRT says that there exists a $y_f$ for every $f \in \mathcal H^*$ such that for $y \in \mathcal H$, we have $f(y) = \langle y,y_f\rangle$. With this established, we define $$ \langle f,g \rangle := \langle y_f,y_g\rangle. $$

We claim that $\Phi(y_{f_{\varphi}}) = \varphi$ (that is, $y_{f_{\varphi}}$ is "the $y$ that we're looking for"). Indeed, we note that for any $f \in \mathcal H^*$, we have $$ [\Phi(y_{f_{\varphi}})](f) = f(y_{f_{\varphi}}) = \langle y_{f_\varphi},y_f \rangle = \langle f, f_{\varphi}\rangle = \varphi(f) $$

  • $\begingroup$ ..@Ben Grossmann I have some confusion in ur third paragraph that we have $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle$. I think it should be $f(y_{f_{\varphi}}) = \langle y_{f_{\varphi}}, f_{\varphi}\rangle$. My doubts is that why written $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle?$. $\endgroup$
    – jasmine
    Commented Jan 13, 2021 at 11:29
  • $\begingroup$ @jasmine It is not clear why you think that. Note that your equation $f(y_{f_{\varphi}}) = \langle y_{f_{\varphi}}, f_{\varphi}\rangle$ implies that $f(y_{f_{\varphi}})$ is the same for every $f \in \mathcal H^*$. $\endgroup$ Commented Jan 13, 2021 at 13:39
  • $\begingroup$ @jasmine Do you understand the Riesz Representation Theorem (which I refer to as "RRT")? $\endgroup$ Commented Jan 13, 2021 at 13:43
  • $\begingroup$ yes i understand @Benn Grossmann and My thinking : RRT say that "for every $f \in H^*$ there exists a unique $v_f \in H$ such that $f(u)=\langle u,v_f \rangle$ for all $u \in H$....continue next comment $\endgroup$
    – jasmine
    Commented Jan 13, 2021 at 14:39
  • 1
    $\begingroup$ @jasmine Note that $f_{\varphi} \in \mathcal H^*$. If $f \in \mathcal H^*$, then what does $f_\varphi(f)$ mean? $f_{\varphi}(x)$ only makes sense for $x \in \mathcal H$. $\endgroup$ Commented Jan 13, 2021 at 16:29

The point is that $\varphi \in \mathcal{H}^{**}$ was arbitrary, and your proof shows that it agrees with another element of $\mathcal{H}^{**}$ which has a particular form, thus showing that every element of $\mathcal{H}^{**}$ has that special form. To see why that is a proof of reflexivity, let's review the general situation:

Let $X$ be a Banach space. For any $y \in X$, we have a functional $\varphi_y \in X^{**}$ defined by $\varphi_y(f) = f(y)$. The map $y \mapsto \varphi_y$ is always an isometric linear embedding of $X$ into $X^{**}$, so the issue is to check that this map is surjective.

Proving surjectivity is exactly this: Given any $\varphi \in X^{**}$, find a $y \in X$ such that for all $f \in X^*$, $\varphi(f) = f(y)$. The proof you've given here is exactly finding such a $y$.


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