Riesz representation theorem

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[edit] Theorem

Let $f$ be a bounded linear functional in a Hilbert space $X$ over $\mathbb K$ . Then there exists a unique $y\in X$ such that, for all $x\in X$ ,

$f(x)=\langle x,y\rangle$

and

$\lVert f\rVert =\lVert y\rVert$

[edit] Proof

If $f = 0$ take $y = 0$ . Suppose then that $f\neq 0$ and consider $ker f$ , which is a closed space by this proposition. By the Hilbert space decomposition theorem we then have

$X=\ker f\oplus (\ker f)^\perp$

and, since $\ker f \not =X$ , $(\ker f)^\perp$ has an element $z$ such that $\lVert z\rVert =1$ . Note now that, for all $x\in X$

$f(x)z-f(z)x\in \ker f$

and, since $z\in (\ker f)^\perp$ ,

$0=\langle f(x)z-f(z)x,z\rangle =f(x)-\langle x,\overline{f(z)}z\rangle$

that is to say

$f(x)=\langle x,\overline{f(z)}z\rangle =\langle x,y\rangle$

with $y=\overline{f(z)}z$ . To prove uniqueness, suppose that for all $x\in X$

$f(x)=\langle x,y_1\rangle =\langle x,y_2\rangle$

for some two elements $y_1,y_2\in X$ . Then $\langle x,y_1\rangle -\langle x,y_2\rangle=\langle x,y_1-y_2\rangle=0$ for all $x\in X$ . In particular, for $x = y 1 - y 2$ which would imply $\lVert y_1-y_2\rVert ^2=0$ and hence $y 1 = y 2$ . Finally, since the inner product is continuous on the first variable

$f\colon X\to \mathbb K$ , $f(x)=\langle x,y\rangle$

is a linear functional that is continuous, and hence bounded by the bounded operator equivalence theorem. Applying the Cauchy-Schwarz inequality, we get

$|f(x)| =|\langle x,y\rangle|\leq\lVert x\rVert\cdot\lVert y\rVert$

and so $\lVert f\rVert =\sup_{\lVert x\rVert =1}\; |f(x)|\leq \lVert y\rVert$ . If $y = 0$ then $\lVert f\rVert=0$ so $f = 0$ . If not, consider $z=\lVert y\rVert^{-1}y$ to obtain