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,


[edit] Proof

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

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

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

that is to say

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

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 = y1y2 which would imply \lVert y_1-y_2\rVert ^2=0 and hence y1 = y2. Finally, since the inner product is continuous on the first variable

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

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

which yields \lVert f\rVert=\lVert y\rVert.

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