Question

Find second order derivative of an inequality function

Answer 1

Let f
f
be a twice differentiable function on [0,1]
[
0
,
1
]
satisfying f(0)=f(1)=0
f
(
0
)
=
f
(
1
)
=
0
. Additionally ∣∣f″(x)∣∣≤1
|
f
″
(
x
)
|
≤
1
in (0,1)
(
0
,
1
)
. Prove that
∣∣f′(x)∣∣≤12,∀x∈[0,1]
|
f
′
(
x
)
|
≤
1
2
,
∀
x
∈
[
0
,
1
]
The hint we were given is to expand into a first order Taylor polynomial at the minimum of f′
f
′
. So I tried doing that:

As f′
f
′
is differentiable, it is continuous, and attains a minimum at [0,1]
[
0
,
1
]
. Thus we can denote x0
x
0
as the minimum, and expending into a Taylor polynomial of the first order around it gives us, for some c
c
between x
x
and x0
x
0
.

Tx0(x)=f(x0)+f′(x0)(x−x0)+f″(c)2(x−x0)2
T
x
0
(
x
)
=
f
(
x
0
)
+
f
′
(
x
0
)
(
x
−
x
0
)
+
f
″
(
c
)
2
(
x
−
x
0
)
2
Now at x=0
x
=
0
we have
Tx0(0)=f(x0)−x0f′(x0)+x20f″(c)2=0
T
x
0
(
0
)
=
f
(
x
0
)
−
x
0
f
′
(
x
0
)
+
x
0
2
f
″
(
c
)
2
=
0
And at x=1
x
=
1
we have
Tx0(0)=f(x0)+(1−x0)f′(x0)+(1−x0)2f″(c)2=

Answer 2

it's your answer my dear friend