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I'm having trouble filling the steps in this guided proof of Riesz's representation theorem. (I already have a proof I can understand, but I'd like to understand this one too.)

Let $H$ be a Hilbert space, and $\varphi : H \to \mathbb{C}$ a bounded linear functional. If $\varphi = 0$ then we are done; otherwise, by scaling, we may assume without loss of generality that $\| \varphi \| = 1$. So, for each $n$, there is a unit vector $h_n$ in $H$ such that $| \varphi(h_n) | > 1 - \frac{1}{n}$. By multiplying each one by an appropriate complex number of unit modulus, we may assume $\varphi(h_n) \in \mathbb{R}$ and $\varphi(h_n) > 1 - \frac{1}{n}$.

Now I run into a problem — I can see that everything follows from the first step, but the first step is eluding me at the moment.

  1. $h_n \longrightarrow h$ for some $h$ in $H$. Why? If $H$ is finite-dimensional then certainly there is a convergent subsequence, but I don't see how we can assert the existence of such an $h$ without knowing more about the relative distances of the $h_n$.

  2. $h$ is orthogonal to the kernel of $\varphi$. I think the idea here is to show that $\| h - u \|$ is minimised over $u \in \ker \varphi$ when $u = 0$, by exploiting the fact that $\| h - u \| \ge | \varphi(h - u) | = | \varphi(h) | = 1$.

  3. $\ker \varphi \oplus \operatorname{span} \{ h \} = H$, by e.g. rank–nullity or orthogonal decomposition.

  4. Hence $\varphi(x) = \langle x, h \rangle$ for all $x \in H$, by decomposing $x$ using the above decomposition of $H$ and linearity.

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Note that for $n,m \geq N$ we have $\varphi(h_n + h_m) \gt 2 (1-\frac{1}{N})$ (assuming that $h_n$ is modified by multiplying with an appropriate scalar as you indicated). By the parallelogram law we then have $$4 = 2\|h_n\|^2 + 2\|h_m\|^2 = \|h_n + h_m\|^2 + \|h_n - h_m\|^2 \geq (\varphi(h_n + h_m))^2 + \|h_n - h_m\|^2$$

Thus, $\|h_n - h_m\|^2 \lt 4 - 4 (1-\frac{1}{N})^2$ and we see that $(h_n)$ is Cauchy.

Added. I agree with Mark's assessment that the outlined proof is a bit convoluted (it needs some useful techniques in Hilbert space geometry, though).

The full version of the Riesz representation theorem can be proved in a few lines:

The map $\Phi: H \to H^{\ast}$ given by $y \mapsto \langle \cdot, y\rangle$ is a conjugate linear isometric isomorphism.

By Cauchy-Schwarz $\|\Phi(y)\| \leq \|y\|$. Since $\|y\|^2 = \langle y,y \rangle = [\Phi(y)](y)$ we have equality, hence $\Phi$ is isometric. Thus, the only point that deserves elaboration is the fact that $\Phi$ is onto. If $\varphi \neq 0$, choose $y' \perp \ker{\varphi}$ with $\varphi(y') = 1$ (this $y'$ exists and is unique because of the orthogonal decomposition $H = \mathbb{C} \oplus \ker{\varphi}$, as $\varphi \neq 0$). Clearly $x - \varphi(x)y' \in \ker{\varphi}$ and thus the computation $\langle x, y' \rangle = \langle x - \varphi(x)y',y'\rangle + \langle \varphi(x) y', y'\rangle = \varphi(x) \|y'\|^2$ shows that $y = \frac{y'}{\|y'\|^2}$ is the unique $y \in H$ such that $\varphi = \Phi(y)$.

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    $\begingroup$ I just checked in Pedersen's Analysis Now (3.1.9) and remarked that the argument above is suspiciously close to his. I must have memorized it well :) $\endgroup$ Commented Jun 4, 2011 at 18:24
  • $\begingroup$ I guess the argument is convoluted because it tries to mimic the demonstration of the Hilbert projection theorem. en.wikipedia.org/wiki/Hilbert_projection_theorem $\endgroup$ Commented Feb 9, 2025 at 19:33
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Here's another proof of the Riesz representation theorem. This isn't what you asked for, but the proof you give seems terribly intricate to me and not as intuitive as one would like.

We start with some continuous linear functional $\phi \in H^*$. We want to show that $\phi = \phi_y$ for some $y \in H$, where $\phi_y$ is defined by $\phi_y (x) = \langle x,y\rangle $. It is very easy to see that such a $y$ is determined uniquely, so the big question is what is the $y$ that corresponds to $\phi$? If we already knew that $\phi$ was of the above form, could we extract the vector $y$ from it without knowing it a-priori? Well, $y$ will satisfy $\phi(y) \ge 0$ (since this should give the norm of $y$ squared), but this doesn't help us much, as there are definitely many vectors in $H$ for which $\phi$ will give non-negative real values (unless $\phi = 0$, which is a trivial case). So a different approach is needed.

We may try to get our $y$ as the result of evaluating $\phi$ at some point in $H$. But $\phi$ spits out scalars, not vectors! So we can't do that either. Nevertheless, we shouldn't dismiss this approach entirely. Instead, let us think: is there some way we can specify a vector in $H$ by using a scalar, or several scalars? In finite dimension, the answer is an obvious yes: we choose some basis to the space and then every vector is specified by a unique tuple of scalars (its coordinates in the basis). But Hilbert spaces have a concept of basis too - an orthonormal basis! If we fix some orthonormal basis $e_1,e_2,\dots$ of $H$ (let's assume $H$ to be separable, just for the sake of nice notation) then every vector $x$ in $H$ is specified by the sequence of scalars $\langle x,e_n\rangle $, for $x=\sum_{n=1}^{\infty} \langle x,e_n\rangle e_n$. So if we find the scalars $\langle y,e_n\rangle $ for all $n \in \mathbb{N}$, we also find $y$. But given the functional $\phi_y$ these scalars can be computed by its very definition, just by evaluating it at the basis elements (and conjugating).

Specifically, if we knew that $\phi = \phi_y$ for a specific vector $y \in H$, then we would have $\phi (e_n) = \phi_y (e_n) = \langle e_n,y\rangle =\overline{\langle y,e_n\rangle }$ for all $n \ge 1$, or equivalently $\langle y,e_n\rangle = \overline{\phi(e_n)}$ for all $n$. But this expresses the scalars $\langle y,e_n\rangle $ in a way which depends only on $\phi$, so by uniqueness, the coordinates of the $y$ which we are looking for are simply $\overline{\phi(e_n)}$! Thus the required $y$ is the vector $\sum_{n=1}^{\infty} \overline{\phi(e_n)}e_n$. It only remains to show that this vector indeed satisfies $\phi = \phi_y$. Since $e_1,e_2,\dots$ is an orthonormal basis and $\phi$ is a continuous linear functional, it suffices to show that both functionals give the same value on the basis elements. And indeed, for every $m$, $\phi_y (e_m) = \langle e_m,y\rangle = \sum_{n=1}^{\infty} \langle e_m,\overline{\phi(e_n)}e_n\rangle = \sum_{n=1}^{\infty} \phi(e_n)\langle e_m,e_n\rangle = \phi(e_m)$ by orthogonality.

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    $\begingroup$ @Mark: One can actually do without using a basis whatsoever (a proof is in Rudin's 'Real and Complex Analysis' for example): Let $M = \{x| \phi(x) = 0\}$. If $\phi\ne 0$, then there is a $z \perp M$ with $\|z\| = 1$. But then $\phi(x) = \phi(x)\langle z , z\rangle = \langle \phi(x) z, z\rangle = \langle \phi(z)x, z\rangle$ (note that $\phi(x)z - \phi(z)x \in M$). Therefore $y = \overline{\phi(z)}z$ is our choice. $\endgroup$ Commented Jun 4, 2011 at 15:50
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    $\begingroup$ I gave +1 for the effort and the good motivation. But I agree with @Sam that the base-free proof is much shorter, simpler and cleaner. I'd add that this proof makes the Riesz representation theorem geometrically obvious. Also, for me it is not quite clear from your exposition why the a priori formal sum $\sum \overline{\phi(e_n)}e_n$ actually belongs to $H$. $\endgroup$ Commented Jun 4, 2011 at 16:58
  • $\begingroup$ I think both proofs are about the same length and complexity ("my" proof can be written in just a few lines, I simply tried to motivate it a bit). As for the base thingy, well, here you have to choose an orthobasis, while in "your" proof you have to choose some vector. So both proofs require some arbitrary choice, but many Hilbert spaces come a-priori with a natural orthobase, and then the above proof gives a recipe for finding the vector which represents $\phi$ in a fairly natural way. $\endgroup$ Commented Jun 4, 2011 at 18:12
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    $\begingroup$ @RenanManeliMezabarba Here's one way: We can write $H = \ker \phi \oplus (\ker \phi)^\bot$. The latter subspace is isomorphic to the image of $\phi$ and so (assuming that $\phi \ne 0$) has dimension 1. If $\left\{ z\right\} $ is a basis for that subspace then $\phi (e_n) = \left\langle e_{n},z\right\rangle \phi\left(z\right)$. The convergence then follows from Bessel's inequality. As you can see, this argument is very similar to the standard proof of RIesz's theorem, so I suppose it doesn't simplify much except in the finite dimensional case (because then there is no question of convergence). $\endgroup$ Commented May 20, 2021 at 14:19
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    $\begingroup$ @Mark How to prove that the orthogonal complement of null(kernel) space is isomorphic to the image? $\endgroup$ Commented Dec 6, 2023 at 15:25
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As @Mark and @t.b. note, this proof is a bit convoluted, but only because of how it is presented.

It would be much more enlightening to note fact 3 first: $$H\cong \ker(\phi)\oplus \text{im}(\phi)$$ Since the image of $\phi$ is a 1 dimensional space, we immediately know that we only need to look for a single vector, which has been denoted $h$ above. The limit construction is just bells and whistles to specify $h$.

We may restrict to the unit circle, where every vector is of the form: $$v = u + ch$$ where $u$ is orthogonal to $h$ and $c^2+|u|^2=1$. From this, $v = h$ maximizes $\phi(v)$ on the unit circle, since $\phi(u)=0$ so we wish to minimize $u$. WLOG if we assume $\phi$ is normalized: $$\lVert \phi\rVert :=\limsup_{\lVert x \rVert = 1} \phi(x)= 1$$ Then fortuitously we have that the following quantities are numerically coincident: $$|\phi(h)| = 1 = \langle h,h\rangle$$ $$|\phi(u)| = 0 = \langle u,h\rangle$$ Combining these using linearity, we prove the standard Riesz representation theorem:

For every continuous (=bounded) linear functional $\phi$, there exists a unique vector $h$ such that: $$\phi(x)=\langle x,h\rangle$$

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