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
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
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 .