Projection and Riesz Representation

The geometry of an inner product and the completeness of L-squared combine into two theorems that nothing in Hilbert space theory does without. The first says one can always project onto a closed convex set, landing on a unique nearest point; for a subspace this is the orthogonal projection. The second says every continuous linear functional on the space is an inner product against a fixed vector, which is the engine behind the Radon-Nikodym theorem and the existence of adjoints. Both turn on completeness [1]. Throughout, $H$ is a real Hilbert space.

#The projection theorem

A set $C$ is convex when it contains the segment between any two of its points, that is $tx+(1-t)y\in C$ for all $t\in[0,1]$ whenever $x,y\in C$ . The proof below uses only the midpoint $\tfrac12(x+y)\in C$ . Closedness and convexity together force a unique nearest point.

Theorem1

Let $C\subseteq H$ be nonempty, closed, and convex. For every $x\in H$ there is a unique $p\in C$ minimising $\norm{x-c}$ over $c\in C$ .

Proof

Let $d=\inf_{c\in C}\norm{x-c}$ and take a sequence $c_n\in C$ with $\norm{x-c_n}\to d$ . Apply the parallelogram law to $x-c_n$ and $x-c_m$ ,

\norm{c_n-c_m}^2=2\norm{x-c_n}^2+2\norm{x-c_m}^2-4\Big\|x-\tfrac{c_n+c_m}{2}\Big\|^2. \tag{1}

The midpoint $\tfrac12(c_n+c_m)$ lies in $C$ by convexity, so its distance from $x$ is at least $d$ , and the last term is at most $-4d^2$ , giving $0\le\norm{c_n-c_m}^2\le 2\norm{x-c_n}^2+2\norm{x-c_m}^2-4d^2$ . As $n,m\to\infty$ the upper bound tends to $2d^2+2d^2-4d^2=0$ , so $\limsup_{n,m}\norm{c_n-c_m}^2\le 0$ , which with $\norm{c_n-c_m}^2\ge 0$ forces $\norm{c_n-c_m}^2\to 0$ . Hence $(c_n)$ is Cauchy and, by completeness, converges to some $p$ , which lies in $C$ because $C$ is closed, with $\norm{x-p}=\lim\norm{x-c_n}=d$ . If $p'$ is another minimiser, the same identity applied to $x-p$ and $x-p'$ gives $\norm{p-p'}^2\le 2d^2+2d^2-4d^2=0$ , so $p'=p$ .

For a subspace the nearest point has a clean characterisation. The residual $x-p$ is orthogonal to the subspace.

Theorem2

Let $M\subseteq H$ be a closed subspace. The nearest point $p=P_M x$ is the unique element of $M$ with $x-p\perp M$ , and the map $P_M$ is linear, idempotent, and satisfies $\norm{P_M x}\le\norm{x}$ .

Proof

A subspace is convex, so Theorem 1 gives a unique nearest $p$ . For any $m\in M$ and $t\in\R$ the point $p+tm\in M$ , so $\norm{x-p-tm}^2\ge\norm{x-p}^2$ , which expands to $-2t\ip{x-p}{m}+t^2\norm{m}^2\ge 0$ for all $t$ , forcing $\ip{x-p}{m}=0$ . Thus $x-p\perp M$ . Conversely if $q\in M$ with $x-q\perp M$ , then for any $m\in M$ , Pythagoras gives $\norm{x-m}^2=\norm{x-q}^2+\norm{q-m}^2\ge\norm{x-q}^2$ , so $q$ is the nearest point and equals $p$ . Linearity follows because $x-P_M x\perp M$ and $y-P_M y\perp M$ give $(\alpha x+\beta y)-(\alpha P_M x +\beta P_M y)\perp M$ with $\alpha P_M x+\beta P_M y\in M$ , so by uniqueness it is $P_M(\alpha x+\beta y)$ . Idempotence is $P_M m=m$ for $m\in M$ , and Pythagoras on $x=P_M x+(x-P_M x)$ gives $\norm{x}^2=\norm{P_M x}^2+\norm{x-P_M x}^2\ge\norm{P_M x}^2$ .

#Orthogonal complement and decomposition

The orthogonal complement of a set $S$ is $S^\perp=\{y\in H:\ip{y}{s}=0\text{ for all }s\in S\}$ , always a closed subspace because it is an intersection of kernels of the continuous functionals $\ip{\cdot}{s}$ . Projection splits the space along it.

Corollary3

For a closed subspace $M$ , every $x\in H$ has a unique decomposition $x=u+v$ with $u\in M$ and $v\in M^\perp$ , namely $u=P_M x$ and $v=x-P_M x$ . Thus $H=M\oplus M^\perp$ , and $(M^\perp)^\perp=M$ .

Proof

The decomposition $x=P_M x+(x-P_M x)$ has $P_M x\in M$ and $x-P_M x\in M^\perp$ by Theorem 2. If $x=u+v=u'+v'$ are two such, then $u-u'=v'-v$ lies in $M\cap M^\perp$ , where a vector is orthogonal to itself and hence zero, so the decomposition is unique. The inclusion $M\subseteq(M^\perp)^\perp$ is immediate, and if $x\in(M^\perp)^\perp$ then $x-P_M x\in M^\perp$ by Theorem 2, while $x-P_M x$ also lies in $(M^\perp)^\perp$ because both $x$ and $P_M x\in M\subseteq(M^\perp)^\perp$ do, so it is orthogonal to itself and vanishes. Hence $x=P_M x\in M$ , giving the reverse inclusion.

#Riesz representation

A linear functional $\varphi:H\to\R$ is bounded when $\abs{\varphi(x)}\le\norm{\varphi}\,\norm{x}$ for a finite constant $\norm{\varphi}$ , equivalently continuous. Every such functional is an inner product in disguise.

Theorem4

For every bounded linear functional $\varphi$ on $H$ there is a unique $y\in H$ with $\varphi(x)=\ip{x}{y}$ for all $x$ , and $\norm{\varphi}=\norm{y}$ .

Proof

If $\varphi=0$ take $y=0$ . Otherwise the kernel $N=\{x:\varphi(x)=0\}$ is a closed proper subspace, so by Corollary 3 the complement $N^\perp$ contains a nonzero vector, which we scale to a unit vector $z$ . For any $x$ , the vector $x-\dfrac{\varphi(x)}{\varphi(z)}\,z$ lies in $N$ , because $\varphi$ sends it to zero, so it is orthogonal to $z$ , giving $\ip{x}{z}=\dfrac{\varphi(x)}{\varphi(z)}\ip{z}{z} =\dfrac{\varphi(x)}{\varphi(z)}$ . Hence $\varphi(x)=\varphi(z)\ip{x}{z}=\ip{x}{\varphi(z)z}$ , so $y=\varphi(z)z$ represents $\varphi$ . Uniqueness follows because $\ip{x}{y-y'}=0$ for all $x$ forces $y=y'$ on taking $x=y-y'$ . For the norm, Cauchy-Schwarz gives $\abs{\varphi(x)}=\abs{\ip{x}{y}}\le \norm{y}\norm{x}$ so $\norm{\varphi}\le\norm{y}$ , while $\varphi(y)=\norm{y}^2$ gives the reverse, so $\norm{\varphi}=\norm{y}$ .

These two theorems generate conditional expectation, the Radon-Nikodym density, and operator adjoints. The orthogonal projection is exactly conditional expectation, the projection of a random variable onto the closed subspace of variables measurable with respect to the conditioning information, and it is the least-squares solution of an overdetermined system. The Riesz representation theorem is the existence half of the Radon-Nikodym theorem, where a density is produced as the vector representing an absolutely continuous functional. It also gives every bounded operator an adjoint, the construction the spectral theorem needs. Completeness was the one nonformal ingredient in both, the property that let the nearest point exist, which is why the Hilbert space, and not merely the inner product space, is the right setting.

[1]

W. Rudin, Functional Analysis, 2nd ed. McGraw-Hill, 1991.

Explore connections

see in the atlas

referenced by (2)

cite

@misc{projection-and-riesz,
  author = {Zac Kienzle},
  title  = {Projection and Riesz Representation},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/projection-and-riesz}
}

#The projection theorem

Theorem1

Let $C\subseteq H$ be nonempty, closed, and convex. For every $x\in H$ there is a unique $p\in C$ minimising $\norm{x-c}$ over $c\in C$ .

Proof

Let $d=\inf_{c\in C}\norm{x-c}$ and take a sequence $c_n\in C$ with $\norm{x-c_n}\to d$ . Apply the parallelogram law to $x-c_n$ and $x-c_m$ ,

\norm{c_n-c_m}^2=2\norm{x-c_n}^2+2\norm{x-c_m}^2-4\Big\|x-\tfrac{c_n+c_m}{2}\Big\|^2. \tag{1}

For a subspace the nearest point has a clean characterisation. The residual $x-p$ is orthogonal to the subspace.

Theorem2

Proof

#Orthogonal complement and decomposition

Corollary3

Proof

#Riesz representation

Theorem4

For every bounded linear functional $\varphi$ on $H$ there is a unique $y\in H$ with $\varphi(x)=\ip{x}{y}$ for all $x$ , and $\norm{\varphi}=\norm{y}$ .

Proof

[1]

W. Rudin, Functional Analysis, 2nd ed. McGraw-Hill, 1991.

Explore connections

see in the atlas

referenced by (2)

cite

@misc{projection-and-riesz,
  author = {Zac Kienzle},
  title  = {Projection and Riesz Representation},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/projection-and-riesz}
}