Let f be a bounded linear functional in a Hilbert space X over . Then there exists a unique such that, for all ,
If f = 0 take y = 0. Suppose then that and consider kerf, 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
To prove uniqueness, suppose that for all
for some two elements .
Then for all . In particular, for x = y1 − y2 which would imply 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 . If y = 0 then so f = 0. If not, consider to obtain
which yields .
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