Rolle's theorem and the intermediate value theorem [Bearbeiten]
Exercise
Is there a
such that
has two distinct zeros in
?
Solution
Suppose there is a
such that
has two zeros
. Since
is continuous on
and differentiable on
, according to Rolle's theorem there is a
with
. But now there is for every
:
. So
cannot have two zeros.
Proof
is continuous on
and
. Therefore, according to the intermediate value theorem, there is an
with
.
Further,
is continuous on
, differentiable on
and
. According to Rolle's theorem, there is hence a
with
. So
has at least one zero (on
).
Exercise (Useful inequality 1)
Show that: For all
there is
Solution (Useful inequality 1)
Proof step: 
Fall 1: 
We define
by
. Then
is continuous, and on
differentiable. Thus the mean value theorem is applicable, and there exists a
with
Now, because of
,
Fall 2: 
Here we have
, i.e., equality.
Fall 3: 
Again, by the mean-value theorem there is a
with
So there is
for all
Proof step: 
Here we show only the case
:
We again define
. Then
is continuous, and on
differentiable. According to the mean value theorem, there exists a
with
For
there is again equality, and for
the statement follows analogously with the mean value theorem.
Hint
From the first inequality, the transition to
for
still allows for the inequality:
Exercise (Useful inequality 2)
Show that: For
there is
Solution (Useful inequality 2)
Proof step: 
Let
. Then the sine function on
is continuous and on
it is differentiable. With the mean value theorem there is a
with
But now there is
for
. Thus we have
Proof step: 
Let
. Then the tangent function on
is continuous and on
it is differentiable. With the mean value theorem there is a
with
But now there is
for
. Thus we have
Hint
The inequality can be further extended to all
:
Where equality only holds at
.
Exercise (Implication of the mean value theorem)
Show by the mean value theorem that:
Let
be continuous on
and differentiable on
. Furthermore, let
and
hold on
. Then, there is
on
.
As an application: prove the following generalization of the Bernoulli inequality: For
and all
there is
.
Proof (Implication of the mean value theorem)
We consider the auxiliary function
This function is continuous and differentiable on
. Further there is

for all 
1st way: By the mean value theorem
By the mean value theorem, for all
there is a
with

for all
2nd way: By the monotony criterion
It follows from 1 that
is monotonically increasing on
(even on
).

for all
![{\displaystyle x\in [a,b]\Leftrightarrow f(x)\geq g(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/564b2103c1390b82a5ff2829946718bae654cfb6)
for all
3rd way: By the fundamental theorem of calculus
By assumption,
is integrable and because of the monotonicity of the integral there is for all
:
But now, by the fundamental theorem of calculus
Concerning the application exercise: We define
and
Then
and
are continuous on
and differentiable on
with
and
Furthermore, there is
. Since the exponential function is strictly monotonically increasing, there is for all
:
With the proven statement we hence get for all
:
Hint
The generalized Bernoulli inequality can even be shown for all
. Equality only holds in the case
.
Solution (mean valuer theorem for continuously differentiable functions)
We consider any function
with the given properties and the secant through the points
and
. The slope of the secant is given by the difference quotient
. Next we look at the slope of the graph, i.e. the derivative values of the function on the interval
.
Fall 1: The function graph is a straight line.
Then the derivative function is constant and consequently there is
for all
.
Fall 2: The function graph is not a straight line.
Then a
must exist with
or
so that
has no straight line as a function graph. It follows in turn that a
exists with
or
, because otherwise the graph at
can never take the function value
. So in total there exist
with
. According to the intermediate value theorem, which is applicable here to the derivative function because it is continuous, there is now a
with
.
So in every m case there is a
with
.
Exercise (Application of the second mean value theorem)
Let
be differentiable. Further let
for all
. Show that then also
holds for all
.
Proof (Application of the second mean value theorem)
Let
be arbitrary with
. Then
and
are continuous on
and differentiable on
according to the assumption. Then, with the second mean value theorem, there is a
with
Since by assumption
holds for all
, we get
.
From this we obtain
Lipschitz-continuity of functions[Bearbeiten]
Exercise (Lipschitz-continuity of functions)
Show by the mean value theorem (using the implication function about Lipschitz continuous functions), that the following functions are Lipschitz continuous. Determine in addition some suitable Lipschitz constants.



Solution (Lipschitz-continuity of functions)
Part 1: For all
there is
So
has a bounded derivative, and is therefore Lipschitz continuous. Further for all
there is
Therefore,
is an appropriate Lipschitz constant.
Part 2: Here, for all
there is:
Thus, by boundedness of derivative,
is Lipschitz-continuous, as well. Moreover, there is for all
Hence,
is an appropriate Lipschitz constant here.
Part 3: finally there is for all
:
So also
is Lipschitz-continuous, and for all
there is
Hence,
is an appropriate Lipschitz constant.
Exercise (Lipschitz-continuity of functions)
Let
be a continuous function that is differentiable on
. Let
be any two real numbers such that
for all
. Prove that for all
the following estimate holds:
Proof (Local Lipschitz continuity)
Let
be a continuously differentiable function, where
is open. Let
be arbitrary. Since
is open, there is a
such that
. Choose
. Then
. Since
is continuous,
is a bounded function on
.
Now we have already proved that differentiable functions with bounded derivatives are Lipschitz continuous. Thus
is Lipschitz continuous on
and thus also on
.