The Riesz representation theorem identifies Hilbert-space duality with the space itself. This is the reason Hilbert spaces support adjoints, variational formulations, and weak equations so cleanly.
Hilbert-space duality
Riesz representation is the point where Hilbert spaces differ sharply from general Banach spaces: every continuous functional is an inner product with a vector.
Theorem. For every \(f\in H^*\), there is a unique \(y\in H\) such that \(f(x)=\langle x,y\rangle\) for all \(x\in H\), and \(\|f\|=\|y\|\).
Proof. If \(f=0\), take \(y=0\). Otherwise, \(\ker f\) is a closed proper subspace. Choose \(z\in(\ker f)^\perp\) with \(\|z\|=1\). Since \(z\notin\ker f\), \(f(z)\ne0\). For any \(x\), the vector \(x-(f(x)/f(z))z\) lies in \(\ker f\), so it is orthogonal to \(z\). Thus \(\langle x,z\rangle=(f(x)/f(z))\langle z,z\rangle=f(x)/f(z)\). Hence \(f(x)=\langle x,\overline{f(z)}\,z\rangle\) with the stated convention.
Uniqueness follows because if \(\langle x,y_1-y_2\rangle=0\) for all \(x\), choosing \(x=y_1-y_2\) gives zero norm. Cauchy-Schwarz gives \(\|f\|\le\|y\|\), and equality follows by evaluating at \(x=y/\|y\|\) when \(y\ne0\).
Riesz map
The Riesz map is conjugate-linear or linear depending on convention, but its content is invariant: it identifies geometry with duality.
Proposition. The map \(R:H\to H^*\), \(Ry(x)=\langle x,y\rangle\), is a conjugate-linear surjective isometry under the convention that the inner product is linear in the first variable.
Proof. Conjugate linearity in \(y\) follows from conjugate-linearity of the second slot. The previous theorem gives surjectivity and norm preservation.
Adjoints via Riesz
Hilbert adjoints are Banach adjoints transported through the Riesz map. This is why adjoints in Hilbert space have such concrete formulas.
Proposition. If \(T\in\mathcal L(H,K)\) between Hilbert spaces, there is a unique \(T^*\in\mathcal L(K,H)\) such that \(\langle Tx,y\rangle_K=\langle x,T^*y\rangle_H\) for all \(x\in H,y\in K\). Moreover \(\|T^*\|=\|T\|\).
Proof. For fixed \(y\in K\), the map \(x\mapsto\langle Tx,y\rangle\) is a bounded linear functional on \(H\). Riesz gives a unique vector \(T^*y\) representing it. Linearity of \(T^*\) follows from uniqueness of representing vectors. The estimate \(\|T^*y\|\le\|T\|\|y\|\) gives boundedness. Applying the same argument to \(T^*\) and using \(\|T\|^2=\|T^*T\|\le\|T^*\|\|T\|\) gives equality, or use the Banach-space adjoint norm equality.
Lax-Milgram theorem
Lax-Milgram is Riesz representation for coercive bilinear forms. It turns weak variational problems into existence and uniqueness statements.
Theorem. Let \(a:H\times H\to\mathbb F\) be a bounded sesquilinear form, linear in the first variable, and coercive: \(\operatorname{Re}a(x,x)\ge c\|x\|^2\) for some \(c>0\). For every \(f\in H^*\), there is a unique \(u\in H\) such that \(a(u,v)=f(v)\) for all \(v\in H\).
Proof. By Riesz, for each \(u\) there is a unique \(Au\in H\) such that \(a(u,v)=\langle Au,v\rangle\). Boundedness of \(a\) makes \(A\in\mathcal L(H)\). Coercivity gives \(c\|u\|^2\le |a(u,u)|=|\langle Au,u\rangle|\le\|Au\|\|u\|\), hence \(\|Au\|\ge c\|u\|\); in particular \(A\) is injective and has closed range.
If \(w\perp\operatorname{Ran}A\), then \(0=\langle Au,w\rangle=a(u,w)\) for all \(u\). Taking \(u=w\) gives \(0=a(w,w)\), and coercivity gives \(w=0\). Thus \(\operatorname{Ran}A\) is dense; since it is also closed, it is all of \(H\). For \(f(v)=\langle y,v\rangle\), solve \(Au=y\). Uniqueness follows from injectivity.
Dual of L2
The space \(L^2\) is the canonical infinite-dimensional Hilbert example. Its duality formula is the integral version of the inner product representation.
Example. For \(H=L^2(\Omega,\mu)\), Riesz says every continuous linear functional has the form \(f(g)=\int g\overline h\,\mathrm{d}\mu\) for a unique \(h\in L^2\). This is the Hilbert-space special case of \(L^p\)-duality.