Consider the following Putnam question from the 2018 exam:
Consider a smooth function
such that
, and
and
. Prove that there exists a point
and a positive integer
such that
.
This is a problem from the 2018 Putnam, and only 10 students were able to solve it completely, making it the hardest question on the exam. I spent a day thinking about it, and my “proof” differs a lot from the official solutions, and is really a heuristic.
Proof: Assume that there does not exist any and
such that
. We will compare
with functions of the form
in
. We will prove that
on
. Because
on
as
, we will have proven that
on
and
. Hence,
cannot be smooth.
Why is ? Let us first analyze what
looks like. It is easy to see that
for
. This is because as
, if
for
, when
will have to decrease to
at
. Hence, there will be a negative derivative involved, which is a contradiction. Hence,
for
, and by continuity of derivatives for smooth functions, all derivatives at
are also
.
Now consider the functions , which are
at
and
at
. These are the same endpoints for
in
. If
ever crosses
in
, then it will have a higher
th derivative than
at the point of intersection. As its
th derivative is also non-negative,
will just keep shooting above
, and hence never “return” to
at
. This contradicts the fact that
. Hence,
will always be bounded above by
in
. As this is true for all
,
on
and
. This contradicts the fact that
is continuous.