# Riesz representation theorem

## [edit] Theorem

Let *f* be a bounded linear functional in a Hilbert space *X* over . Then there exists a unique such that, for all ,

and

## [edit] Proof

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

and, since , has an element *z* such that . Note now that, for all

and, since ,

that is to say

with . To prove uniqueness, suppose that for all

for some two elements .
Then for all . In particular, for *x* = *y*_{1} − *y*_{2} which would imply and hence *y*_{1} = *y*_{2}.
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 . If *y* = 0 then so *f* = 0. If not, consider to obtain

which yields .

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