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We first prove the following lemma.
If
is a continuous function with
then there exists a
such that .
Define the sequences and inductively, as follows.
We note that
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(1) |
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(2) |
By the fundamental axiom of analysis
and
. But
so
. By continuity of
But we have
and
so that
. Furthermore we have
, proving the assertion.
Set
where
. satisfies the same conditions as before, so there exists a such that . Thus proving the more general result.
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