Page 61 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes and using the inequality
1
q q
(1 t ) 1, t (0, 1],
We immediately get f (t) < f (0), t (0, 1],
and the Lemma again follows.

For the remainder of the proof we are going to assume that u, v > 0.
Obviously f is differentiable on (0, 1) and the solutions of the equation (3)
f (t) = 0
Let s be defined as in (1), so under the assumption that u, v > 0, we clearly have 0 < s < 1.

We are going to prove first that s is the unique solution in (0, 1) of the equation (3).
We have

1
1 q q 1 q 1
f (t) = u v (1 t ) q t … (4)
q
t q
= u v , t (0, 1)
1 t q
so the equation (3) reads

t q
u v = 0.
1 t q
Equivalently, we have

1
t q p
= u/v,
1 t q
t q
p
= (u/v) ,
1 t q
(u/v) p u p
2
t = ,
1 (u/v) p u p v p
Having shown that the “candidates” for the maximum point are 0, 1 and s let us show that s is the
only maximum point.
For this purpose, we go back to (4) and we observe that f is also continuous on (0, 1).

Since lim f (t) = u > 0 and
t 0
lim f (t) = –
t 1

and the equation (3) has exactly one solution in (0, 1), namely s, this forces
f (t) > 0 t (0, s)

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