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### The Riesz Representation Theorem
The Riesz representation theorem gives us a characterization of the set of all linear functionals on a finite dimensional inner product space. The advantage of this theorem is that it gives an expression for a whole class of linear functionals. These ideas motivated many other results in functional analysis and in advanced analysis. We start by defining linear functionals.
**Definition 61**.: A _linear functional_\(f\) on a vector space \(X\) is a linear map \(X\to\mathbb{F}\) where \(\mathbb{F}\) is the scalar field of \(X\). If \(X\) is a real inner product space then \(\mathbb{F}=\mathbb{R}\), and \(\mathbb{F}=\mathbb{C}\) if \(X\) is complex.
**Definition 62**.: A linear functional on \(X\) is said to be _bounded_ if there exists a constant \(M\geq 0\) such that
\[|f(x)|\leq M||x||\]
for all \(x\in X\).
For a given bounded linear functional \(f\), we may find many \(M\)’s satisfying the above condition. If it is valid for one \(M\), it is valid for \(M_{1}>M\) also. The next theorem shows that the condition of boundedness is equivalent to continuity. For the proof of the following theorem we refer to, e.g., [35], p. 97.
**Theorem 75**.: _A linear functional \(f\) on \(X\) is continuous if and only if \(f\) is bounded._
In the following we list some examples of linear functionals:
* \(f:\mathbb{R}^{n}\to\mathbb{R}\), where \(f(x)=\sum_{k=1}^{n}a_{k}x_{k}\), \(x=(x_{1},x_{2},\ldots,x_{n})\in\mathbb{R}^{n}\) and \((a_{1},a_{2},\ldots,a_{n})\in\mathbb{R}^{n}\) is fixed.
* \(f:C[a,b]\to\mathbb{R}\), where \(f(x)=\int_{a}^{b}x(t)\,dt\), \(x\in C[a,b]\).
Note that for a normed space \(X\) and \(x\in X\), \(f:X\to\mathbb{R}\) defined as \(f(x)=||x||\) is a functional on \(X\) which is not linear.
Suppose \(X\) is a finite dimensional, nonzero inner product space over \(\mathbb{F}\). Let \(v\in X\) be a fixed vector, then for \(x\in X\) the mapping \(f\) given by:
\[f(x)=\langle x,v\rangle\]
is a linear functional. Linearity follows from
\[f(a_{1}x_{1}+a_{2}x_{2}) = \langle a_{1}x_{1}+a_{2}x_{2},v\rangle\] \[= a_{1}\langle x_{1},v\rangle+a_{2}\langle x_{2},v\rangle\] \[= a_{1}f(x_{1})+a_{2}f(x_{2}).\]
The Riesz theorem states that there are in fact no other types of linear functionals on a finite dimensional inner product space, thus any linear functional on the space \(X\) must have the form \(f(x)=\langle x,w\rangle\) for some unique point \(w\) in \(X\).
**Theorem 76** (The Riesz representation theorem).: _Let \(X\) be a finite dimensional inner product space and let \(f\) be a linear functional on \(X\). Then there exists a unique point \(v\in X\) such that \(f(x)=\langle x,v\rangle\) for all \(x\in X\)._
Proof
: First we show there exists a vector \(v\in X\) such that \(f(x)=\langle x,v\rangle\) for every \(x\in X\). Let \(e_{1},\ldots,e_{n}\) be an orthonormal basis for \(X\). Then,
\[f(x) = f(\langle x,e_{1}\rangle e_{1}+\cdots+\langle x,e_{n}\rangle e_{ n})\] \[= \langle x,e_{1}\rangle f(e_{1})+\cdots+\langle x,e_{n}\rangle f(e _{n})\] \[= \langle x,\overline{f(e_{1})}e_{1}+\cdots+\overline{f(e_{n})}e_{ n}\rangle.\]Setting \(v=\overline{f(e_{1})}e_{1}+\cdots+\overline{f(e_{n})}e_{n}\) we have \(f(x)=\langle x,v\rangle\) for every \(x\in X\) as desired. Now we prove that this vector is unique. Suppose we have two vectors \(v_{1},v_{2}\in X\) such that
\[f(x)=\langle x,v_{1}\rangle=\langle x,v_{2}\rangle\]
for every \(x\in X\). Then
\[0=\langle x,v_{1}\rangle-\langle x,v_{2}\rangle=\langle x,v_{1}-v_{2}\rangle\]
for every \(x\in X\). By taking \(x=v_{1}-v_{2}\) we show that \(v_{1}-v_{2}=0\), in other words \(v_{1}=v_{2}\).
Suppose \(X\) is finite dimensional and \(f\) is a linear functional on \(X\), then by the above proof the vector \(v\) satisfies \(f(x)=\langle x,v\rangle\) where
\[v=\overline{f(e_{1})}e_{1}+\cdots+\overline{f(e_{n})}e_{n}.\]
The right side of this equation seems to depend on the functional \(f\) as well as the orthonormal basis \(e_{1},e_{2},\ldots,e_{n}\). However the Riesz representation theorem tells us that \(v\) is uniquely determined by \(f\). Hence the right side of the equation is the same regardless of which orthonormal basis is taken for \(X\). The following is a simple consequence of this theorem.
**Theorem 77**.: _All linear functionals on finite dimensional inner product spaces are bounded._
Proof
: If \(f\) is a linear functional on \(X\), then there is a point \(v\in X\) such that \(f\) has the form \(f(x)=\langle x,v\rangle\) for all \(x\in X\). But then by the CSB inequality
\[|f(x)|=|\langle x,v\rangle|\leq||x||||v||,\]
so \(f\) is bounded.
Note that we can say even more; if we set \(||f||=\sup_{||x||=1}|f(x)|\), then \(||f||\leq||v||\). In fact \(||f||=||v||\). This is seen by noting that
\[|f(v)|=|\langle v,v\rangle|=||v||^{2}\]
so that we cannot have \(|f(x)|<||x||||v||\) for all \(x\in X\).
**Remark 41**.: There are many other versions of the Riesz representation theorem giving similar results in other spaces. Even though we proved this theorem for finite dimensional inner product spaces, for a complete inner product space called a _Hilbert space_, the Riesz representation theorem gives rise to the quite important notion of an operator _adjoint_ to a given operator.
**Theorem 78**.: _If \(f\) is a bounded linear functional on a Hilbert space \(H\), then there is a unique point \(v\in H\) such that \(f(x)=\langle x,v\rangle\) for all \(x\in H\)._
The proof of this theorem is considerably more involved (for proof consult [35], pp. 188-190), but note that here we require the functional to be bounded, whereas in the former case boundedness is obtained as a consequence.
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**Example 73**.: Each of the following defines a norm on \(\mathbb{R}^{n}\):
\[||x||_{1} = \sum_{i=1}^{n}|x_{i}|,\] \[||x||_{2} = \left(\sum_{i=1}^{n}|x_{i}|^{2}\right)^{1/2},\] \[||x||_{\infty} = \max_{1\leq i\leq n}|x_{i}|.\]
In general, for any \(1\leq p<\infty\), the expression
\[||x||_{p}=\left(\sum_{i=1}^{n}|x_{i}|^{p}\right)^{1/p}\]
defines a norm on \(\mathbb{R}^{n}\).
Note that \(||x||_{2}\) is the usual Euclidean norm.
**Example 74**.: Each of the following defines a norm on \(C[a,b]\)
\[||f||_{1} = \int_{a}^{b}|f(t)|dt,\] \[||f||_{2} = \left(\int_{a}^{b}|f(t)|^{2}dt\right)^{1/2},\] \[||f||_{\infty} = \max_{a\leq t\leq b}|f(t)|.\]
**Example 75**.: If \((X,||\cdot||)\) is a normed vector space and \(Y\) is a linear subspace of \(X\), then \(Y\) is also normed by \(||\cdot||\). In other words, the restriction of \(||\cdot||\) to \(Y\) defines a norm on \(Y\).
**Example 76** (Function Space).: Let \(C[a,b]\) represent the collection of continuous real-valued functions \(f:[a,b]\rightarrow\mathbb{R}\), then
\[||f||:=\sup_{x\in[a,b]}|f(x)|\]
defines a norm on \(C[a,b]\). Note that \(f\) being continuous on the compact set \([a,b]\) implies it is a bounded function by the Extreme Value Theorem. Furthermore, \(||f||=0\) if and only if \(f(x)=0\) for each \(x\in[a,b]\). To show the triangle inequality holds for the above norm observe:
\[||f+g||=\sup_{x\in[a,b]}|f(x)+g(x)|\leq\sup_{x\in[a,b]}|f(x)|+\sup_{x\in[a,b]} |g(x)|=||f||+||g||.\]
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