- Let
be a function with second derivative
continuous and nonzero on an interval
. Furthermore, let
be a constant such that
Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality
- If
prove that
- Proof. Since we have the assumption that
for all
we may divide by
, to obtain
Then, to apply the second mean-value theorem for integrals (Theorem 5.5 of Apostol) we define functions
The function
is continuous since
is a composition of continuous functions (we know
is continuous since it is differentiable) and
is continuous (again, it is differentiable since
exists and is continuous by assumption). Then the product of continuous function is also continuous, which establishes that
is continuous. We also know that
meets the conditions of the theorem since it has derivative given by
This derivative is continuous since
and
are continuous and
is nonzero. Furthermore, this derivative does not change since on
since
is nonzero on
(and by Bolzano’s theorem we know that a continuous function that changes sign must have a zero). Therefore, we can apply the second mean-value theorem:
- Proof. Using part (a), we take
, giving us
Thus,
is continuous and never changes sign. Furthermore,
(where
is a given constant) and
since
. Thus,
Hi, this resource was a great find!
One thing left me stumped working on this. On the second row of the solution to part one, I initially summed the two integrals, getting 2/m rather than 4/m.
I figured that I could use the triangle inequality; but I can’t understand if summing the two integrals gives an incorrect result. Could you explain?
On second thoughts… you can’t actually sum the two integrals when 1/(phi’) is actually evaluated at a and b. But the two integrals can still both be equal to 2.
The second line should be incorrect, but easily recoverable (by splitting the absolute value before this).
A counterexample can possibly be constructed in this way: if two integrals are about to be equal in absolute value but have different signs, say -1 and 1, then the second line evaluates to something around 0, but by choosing the correct phi’, the first line can be farther from 0. Thus the first line > the second line.