Ries Representation Theorem

The Riesz representation theorem (also the Riesz – Fréchet theorem ) is a statement of functional analysis , according to which each linear bounded functional in a Hilbert space can be represented through a scalar product with the help of some element. Named in honor of the Hungarian mathematician Frigyes Rees .

Content

Formulation

Let there be a Hilbert space $H$ ${\ displaystyle H}$ and linear bounded functional $f\in H'$ ${\ displaystyle f \ in H '}$ in space $H$ ${\ displaystyle H}$ . Then there is a single element. $y$ ${\ displaystyle y}$ spaces $H$ ${\ displaystyle H}$ such that for arbitrary $x\in H$ ${\ displaystyle x \ in H}$ performed $f(x)=\langle y,x\rangle$ ${\ displaystyle f (x) = \ langle y, x \ rangle}$ . In addition, the equality holds: $\|y\|=\|f\|$ ${\ displaystyle \ | y \ | = \ | f \ |}$ .

Proof

$\ker(f)$ ${\ displaystyle \ ker (f)}$ the core of a linear functional is a vector subspace $H$ ${\ displaystyle H}$ .

Existence $y$ ${\ displaystyle y}$

If a $f\equiv 0$ ${\ displaystyle f \ equiv 0}$ it is enough to take $y=0$ ${\ displaystyle y = 0}$ . Let's pretend that $f\neq 0$ ${\ displaystyle f \ neq 0}$ . Then $\ker(f)\neq H$ ${\ displaystyle \ ker (f) \ neq H}$ and therefore orthogonal complement $\ker(f)^{\bot }$ ${\ displaystyle \ ker (f) ^ {\ bot}}$ cores $f$ ${\ displaystyle f}$ not equal $\{0\}$ ${\ displaystyle \ {0 \}}$ . Choose an arbitrary nonzero vector $b\in \ker(f)^{\bot }\setminus {\big \{}0{\big \}}$ ${\ displaystyle b \ in \ ker (f) ^ {\ bot} \ setminus {\ big \ {} 0 {\ big \}}}$ . Set $y={\tfrac {f(b)}{\|b\|^{2}}}b$ ${\ displaystyle y = {\ tfrac {f (b)} {\ | b \ | ^ {2}}} b}$ . We will show that $f(x)=\langle y,x\rangle$ ${\ displaystyle f (x) = \ langle y, x \ rangle}$ for all $x\in H$ ${\ displaystyle x \ in H}$ . Consider a vector $p_{x}=x-{\tfrac {f(x)}{f(b)}}b$ ${\ displaystyle p_ {x} = x - {\ tfrac {f (x)} {f (b)}} b}$ . notice, that $f(p_{x})=f(x)-{\tfrac {f(x)}{f(b)}}f(b)=0$ ${\ displaystyle f (p_ {x}) = f (x) - {\ tfrac {f (x)} {f (b)}} f (b) = 0}$ , and thus, $p_{x}\in \ker(f)$ ${\ displaystyle p_ {x} \ in \ ker (f)}$ . Insofar as $b\in \ker(f)^{\bot }$ ${\ displaystyle b \ in \ ker (f) ^ {\ bot}}$ then $\langle b,p_{x}\rangle =0$ ${\ displaystyle \ langle b, p_ {x} \ rangle = 0}$ . Consequently,

$\langle b,p_{x}\rangle ={\Big \langle }b,x-{f(x) \over f(b)}b{\Big \rangle }=\langle b,x\rangle -{f(x) \over f(b)}\|b\|^{2}=0$ {\ displaystyle \ langle b, p_ {x} \ rangle = {\ Big \ langle} b, x- {f (x) \ over f (b)} b {\ Big \ rangle} = \ langle b, x \ rangle - {f (x) \ over f (b)} \ | b \ | ^ {2} = 0} .

From here $f(x)=\langle b,x\rangle {\tfrac {f(b)}{\|b\|^{2}}}$ ${\ displaystyle f (x) = \ langle b, x \ rangle {\ tfrac {f (b)} {\ | b \ | ^ {2}}}$ and $f(x)=\langle y,x\rangle$ ${\ displaystyle f (x) = \ langle y, x \ rangle}$ .

Uniqueness $y$ ${\ displaystyle y}$

Let's pretend that $y$ ${\ displaystyle y}$ and $z$ ${\ displaystyle z}$ items $H$ ${\ displaystyle H}$ satisfy $f(x)=\langle y,x\rangle =\langle z,x\rangle$ ${\ displaystyle f (x) = \ langle y, x \ rangle = \ langle z, x \ rangle}$ .

This means that for all $x\in H$ ${\ displaystyle x \ in H}$ fair equality $\langle y-z,x\rangle =0$ ${\ displaystyle \ langle yz, x \ rangle = 0}$ , in particular $\langle y-z,y-z\rangle =\|y-z\|^{2}=0$ ${\ displaystyle \ langle yz, yz \ rangle = \ | yz \ | ^ {2} = 0}$ where does equality come from $y=z$ ${\ displaystyle y = z}$ .

Equality of norms

For proof $\|y\|=\|f\|$ ${\ displaystyle \ | y \ | = \ | f \ |}$ First, from the Cauchy-Bunyakovsky inequality we have: $|f(x)|=|\langle y,x\rangle |\leq \|y\|\|x\|$ ${\ displaystyle | f (x) | = | \ langle y, x \ rangle | \ leq \ | y \ | \ | x \ |}$ . Hence, according to the definition of the norm of a functional, we have: $\|f\|\leq \|y\|.$ ${\ displaystyle \ | f \ | \ leq \ | y \ |.}$ Besides, $\langle y,y\rangle =f(y)\leq \|y\|\|f\|$ ${\ displaystyle \ langle y, y \ rangle = f (y) \ leq \ | y \ | \ | f \ |}$ from where $\|y\|\leq \|f\|$ ${\ displaystyle \ | y \ | \ leq \ | f \ |}$ . Combining the two inequalities, we get $\|y\|=\|f\|$ ${\ displaystyle \ | y \ | = \ | f \ |}$ .

Ries Representation Theorem

Content

Formulation

Proof

Existence $y$ ${\ displaystyle y}$

Uniqueness $y$ ${\ displaystyle y}$

Equality of norms

See also

Notes

More articles:

Ries Representation Theorem

Content

Formulation

Proof

Existencey {\ displaystyle y}

Uniquenessy {\ displaystyle y}

Equality of norms

See also

Notes

More articles:

Existence $y$ ${\ displaystyle y}$

Uniqueness $y$ ${\ displaystyle y}$