Alternative proof of the Riesz Representation Theorem

There are quite a few different proofs of the theorem. One that I particularly like is the following: Let $f \in \mathscr H'$ be the given functional in the dual of $\mathscr H$. Then define $$\mathcal F[x]:= \frac 12 ||x||^2 - f(x)$$ for $x \in \mathscr H$. One shows that this functional has a minimizer: first, it is bounded below which can be seen using the boundedness of $f$ and the Cauchy-Schwarz or Young inequality. Then, by the boundedness from below, there exists a minimizing sequence. Using the parallelogram identity, one shows that this minimizing sequence is Cauchy, and by completeness it converges. Since $\mathcal F$ is continuous, the limit of the minimizing sequence is a minimizer, let's call it $y_f$.

The first order criterion for minimality is that $\frac{\mathrm d}{\mathrm dt}\vert_{t=0} \mathcal F[y_f + tz]=0$ for any $z \in \mathscr H$. Computing this derivative and plugging in $t=0$ leads to $\langle y_f, z\rangle = f(z)$, which is what we wanted.

The uniqueness of $y_f$ then follows by assuming that there is another such element $\tilde y_f$ and inferring that $\langle y_f - \tilde y_f, z\rangle=0$ for all $z \in \mathscr H$, which implies $y_f = \tilde y_f$.

To see that the map $f \mapsto y_f$ is an isometry (which is also part of the claim of the Riesz representation theorem), one uses $f(x)= \langle y_f, x\rangle$ to first see $||f||_{\mathscr H'} \le ||y_f||$ by Cauchy-Schwarz, and then one plugs in $f(y_f) = ||y_f||^2$ to get the other direction.

This proof is probably not the easiest one to understand when you first encounter the theorem, but it is a very insightful one to me, since it connects to Calculus of Variations and establishes the identity $f(x)= \langle y_f, x\rangle$ as the Euler-Lagrange inequality of a functional - something that comes as a surprise upon seeing it the first time.

I really liked the answer provided by @Lukas and would like to extend it to more general Banach spaces under suitable modifications. Let $X$ be a Banach space and let $J \colon X \to 2^{X^*}$ be its duality mapping given by $$Jx = \{ f \in X^* \colon \langle f,x\rangle = \|x\|^2, \ \|f\|=\|x\| \}.$$ Note that, for every $x\in X$, the set $Jx$ is non-empty by the Hahn-Banach theorem. The Banach space $X$ is called smooth if $Jx=\{j(x)\}$ is a singleton for every $x \in X$. One then identifies the multi-valued map $J$ with the function $j \colon X \to X^*$. In the case where $X$ is a Hilbert space, the duality mapping is just the identity: $j(x)=x$ (understood under the identification $H^*=H$). This turns out to be a characterization of Hilbert spaces.

Theorem A. Let $X$ be a reflexive, smooth, and strictly convex Banach space. Then, for every $f \in X^*$ there exists a unique $ y_f \in X$ with $\|y_f\|=\|f\|$ such that $$ j(y_f) = f.$$

The usual function/sequence spaces $L^p,\ell^p$ ($1<p<\infty$) are reflexive, smooth and strictly convex. You can find the definition of strict convexity here. You should think of it as a geometric property that ensures the uniqueness (but not existence) of minimization problems. The existence requires reflexivity.

The proof of Theorem A is similar: You consider the convex functional $\mathcal F \colon X \to \mathbb R$ given by $$\langle \mathcal F,x \rangle= \frac12\|x\|^2 -\langle f, x\rangle.$$ Then $\mathcal F$ is coercive since $$\langle \mathcal F,x \rangle \ge \frac12\|x\|^2 -\| f\| \|x\| \xrightarrow{|x|\to\infty} \infty.$$ Since $X$ is reflexive, $\mathcal F$ attains its infimum over $X$. To see this, by the coercivity we can pick $r>0$ such that $\langle \mathcal F,x\rangle \ge 1$ for $\|x\| \ge r$. Hence, it suffices to prove that the infimum is attained on the ball $B(0,r)$. We can pick a minimizing sequence $(x_n) \subset B(0,r)$ such that $\langle \mathcal F,x_n\rangle \downarrow \inf_{B(0,r)} \mathcal F$. Since $(x_n)$ is bounded we may assume that $x_n \xrightarrow{w} x_0$. Now the weak (lower semi-)continuity of $\mathcal F$ implies that $\mathcal \langle \mathcal F, x_0\rangle \le \liminf \langle \mathcal F, x_n\rangle$ so that $\inf_{B(0,r)} \mathcal F=\langle \mathcal F, x_0\rangle$. This proves the existence of a minimizer.

The uniqueness of the minimizer follows from the strict convexity of $X$, see here for a similar argument. Moreover, since the (Gateaux) derivative of $\frac12\|x\|^2$ coincides with the duality mapping $j$ (see here and here), the criterion for minimality gives $$j (y_f) = f.$$ In particular, the definition of the duality pairing and the above imply the identity $\|y_f\|= \|j(y_f)\|=\|f\|.$

There is an even more general result which I think is due to Rockafeller.

Theorem B. Let $X$ be a reflexive, smooth, strictly convex Banach space. Let $\phi \colon X \to \mathbb R \cup \{-\infty,\infty\} $ be a proper, convex and lower semicontinuous. Then, for every $f \in X^*$ there exists a unique $y_f \in X$ such that $$ f-j(y_F) \in \partial \phi(y_F). $$ In other words, the multi-valued map $j+\partial \phi$ is a bijection.

Here, $\partial \phi$ denotes the sub-differential of $\phi$. The proof of Theorem B is similar.

Before asking the question, I had previously searched for answers, but nothing that I had found seemed like an actually different proof, just a slight repackaging. I somehow came across one now, and since I asked about it, I might as well share it. It generalises the finite-dimensional Riesz Representation Theorem.

Uniqueness and isometry are as in the proof of Lukas, the difference in this proof is in the existence of the element generating the functional.

Let $H$ be a Hilbert space and $\varphi \in H'$. Let $\mathcal{F}$ denote the set of all finite-dimensional subspaces of $H$. For each $E \in \mathcal{F}$, by the finite-dimensional Riesz representation theorem, there exists a unique vector $y_E \in E$ such that: $$ \varphi(x)=\langle x ; y_E \rangle \qquad \forall x\in E. $$ We obtain our representative vector as the limit of an increasing sequence of such $y_E$.

Let $E,F\in\mathcal{F}$, $E\subseteq F$. Then for any $x \in E$: $$ \langle x; y_F-y_E \rangle = \langle x,y_F\rangle-\langle x,y_E\rangle = \varphi(x)-\varphi(x)=0 $$ Therefore $y_F-y_E\in E^\perp$. Since $y_E\in E$, $y_E \perp (y_F-y_E) $. Therefore: $$ \|y_F\|^2=\|y_E\|^2+\|y_F-y_E\|^2. $$ In other words, $E\subseteq F$ implies $\|y_E\| \le \|y_F\|$.

Also: $$ \|y_E\|= \|\varphi\|_E \le \|\varphi\|. \qquad (1)$$ So the family $\{y_E\}_{E\in\mathcal{F}}$ is bounded (in norm) above by $\|\varphi\|$. Denote its supremum in norm by $\alpha$.

Let $$ E_1\subset E_2\subset E_3\subset \cdots $$ be any increasing sequence in $\mathcal F$. Then $\langle\|y_{E_n}\|^2\rangle_{n}$ is increasing and bounded above, therefore converges.

For $m\le n$, we re-arrange (1): $$ \|y_{E_n}-y_{E_m}\|^2=\|y_{E_n}\|^2-\|y_{E_m}\|^2. $$ The right-hand side vanishes, so the sequence is Cauchy.

Take a sequence $\langle E_n \rangle_{n}$ which approaches $\sup_{E \subseteq \mathcal{F}}\|y_E\|$. Define $F_n = E_1 + \cdots E_n$.

$E_n \subseteq F_n$ so $\|y_{E_n}\| \le \|y_{F_n}\|$; this therefore is also an increasing sequence and thus is Cauchy, and converges to some $y \in H$. Since the sequence is maximising, the norm of $y$ must be $\alpha$.

Given $x \in H$, take $M = \text{span}(x)$, and $G_n = F_n + M$; since $F_n \subseteq G_n$, we can use (1), and the fact $\|y_{G_n}\|^2 \le \alpha^2$: $$ \|y_{G_n} - y_{F_n}\|^2 = \|y_{G_n}\|^2 - \|y_{F_n}\|^2 \le \alpha^2 - \|y_{F_n}\|^2 \to 0 \qquad n \to \infty $$

By definition of $y_{G_n}$: $$ \varphi(x) = \langle x; y_{G_n} \rangle \qquad \forall n \in \mathbb{N}$$ and thus: $$ \varphi(x) = \lim_{n \to \infty}\langle x; y_{G_n} \rangle = \langle x; y \rangle $$ and this holds for any $x \in H$.

This concludes the proof.

The idea of this proof comes entirely from this mathoverflow answer: https://mathoverflow.net/questions/215523/on-the-riesz-representation-theorem