Darboux Theorem statement proof examples and intermediate value property explanation
In Mathematics, the Darboux theorem is a theorem named after Jean Gaston Darboux. It states that every function that is derived from the differentiation of another function has intermediate value property: the image of an interval will also be an interval.
When the function $f$ is continuous differentiable $(f$ in $C^1[(x, y])$, this is the result of the intermediate value theorem. But even when the function $f$ is not continuous, Darboux theorem lays down every limitation on what it can be.
Darboux Theorem Statement
If $f$ is differentiable on $[a,b]$ and if is a number between $f^\prime (a)$ and $f^\prime(b)$, then there is minimum one point $c \in (a,b)$ such that $f^\prime(c)$ is .
This states that the converse of the intermediate theorem is not true because the derivative function of a differentiation function is not certainly continuous. For example,
$ f(t) = \left\{\begin{matrix} t^2 sin\left(\dfrac{1}{t}\right) & \text{for } t \neq 0, \\ 0 & \text{for } t = 0 \end{matrix}\right.$
is continuous and has discontinuous derivatives which satisfy the intermediate value property given by the Darboux theorem.
Darboux Theorem Proof 1
Suppose that $f^\prime(a) < \lambda f^\prime(b)$. Let $F : [a,b] \to$ be defined by $f(x) = f(x)- \lambda x$. Then, $f$ is differentiable on $[a,b]$, because so is the function $f$ by hypothesis. We find $F^\prime(a)= f^\prime(a) - \lambda < 0$ and $F^\prime(b) = f^\prime(b) - \lambda > 0$. Note that $F^\prime(a) < 0$ means $F^\prime(t_1) < F (a)$ for some $t_1 \in (a, b)$. Also for, $F^\prime(b) > 0 $, we can find the value of $t_2(a, b)$ such that $F^\prime(t_2) < F(b)$. Hence, neither $a$ nor $b$ are the points where $F$ can attain absolute minimum. As $F$ is continuous on $[a,b]$, it must attains its relative minimum at some point $c \in (a,b)$ by extreme value theorem for some continuous functions. This implies $F^\prime(c)$ by Fermat's theorem and hence $f^\prime(c) = \lambda$ as desired. This proof follows the similar argument of $f^\prime(b) < \lambda < f^\prime(a)$.
Darboux Theorem Proof 2
Let us assume that $f^\prime(a) < \lambda < f^\prime(b)$ and consider a function $F: [a,b] \to R$ defined by $F(x) = f^\prime(x)- \lambda x$. Then $F^\prime(x) = f^\prime(x)- \lambda$ and determine that $F^\prime(a) = f^\prime(a)- \lambda < 0$ and $F^\prime(b) = f^\prime(b)- \lambda > 0$.This implies that function $F$ is monotonic on $[a,b]$ in the way there exist $x,y,z \in [a,b]$ such that $x < y < z$ satisfying only one the following condition $(1)$ or $(2)$.
$F(x) < F(y)$ and $F(y) >F(z)$. The theorem will be proved for this case.
$F(x) > F(y)$ and $F(y) < F(z)$. Proof in the case will follow a similar argument.
Suppose $(a )$ holds, then we can observe following three cases
$F(x) < F(z)$
$F(z) < F(x)$
$F(z) = F(x)$
Case 1: Suppose that $F(x) > F(z)$. Then $F(x) < F(z) < F(y)$. $F$ is differentiable on $(a,b)$ so is $f$ by hypothesis. Hence $F$ is also continuous on $(a,b)$. Hence, the intermediate value theorem applies to $F$ on $[x,y] \subseteq (a,b)$ and we get $d\subseteq (x,y) \subseteq (a,b)$ such that $F(d) = F(z)$. Note that $a < x < d< y < z < b$.
Now, we can use the Rolle’s theorem to $F$ on the closed interval $(d,z)$ and gets $c \in (d,z)\subseteq (x,z)\subseteq (a,b) $ such that $F^\prime(c) = 0$, which results $f^\prime (c) = \lambda$ as desired.
Case 2: Suppose that $F(z) < F(x)$. Then $F(z) < F(x) < F(y)$. This theorem further follows by argument similar to that of case 1.
Case 3: Suppose that $F(z) = F(x)$. Then, we can finally apply rolle’s theorem for $F$ directly on $[x,z]$ and obtain $c(x,z) \subseteq (a,b)$ such that $F^\prime(c) = 0$ leading to $f^\prime(c) = \lambda$ as desired.
We arrive at the same conclusion if $(2)$ holds by the same argument.
Darboux Problems and Solutions
Consider the function $f(x) = sin (x)$
1. What Values of the Derivatives of $f$ are assured by Darboux Theorem on the Interval $[1,3]$?
Solution:
As $f^\prime (x)= cos x , f^\prime (1)= 0.54030$ and $f^\prime (3)= - 0.98999$. Hence, the assured values are from $- 0.98999$ to $0.54030$.
2. Does Darbous Property Guarantee any Value on the Interval $[0, 2 \pi]$.
No, $f^\prime(x)= 1$ at both the ending points. Hence, there are no values between $1$ and $1$.
FAQs on Darboux Theorem in Real Analysis with Intuition and Applications
1. What is Darboux’s Theorem in calculus?
The Darboux’s Theorem states that the derivative of a function has the Intermediate Value Property, even if the derivative is not continuous. In simple terms, if f′(a) and f′(b) take different values, then f′ takes every value between them for some point in (a, b).
- If f is differentiable on [a, b], then f′ cannot “jump” over any value.
- This means derivatives behave like continuous functions in terms of intermediate values.
- However, the derivative itself does not have to be continuous.
2. What does Darboux’s Theorem say about derivatives?
Darboux’s Theorem says that derivatives satisfy the Intermediate Value Theorem, even when they are not continuous. Specifically:
- If f is differentiable on (a, b),
- And if f′(a) ≠ f′(b),
- Then for any number k between f′(a) and f′(b), there exists c in (a, b) such that f′(c) = k.
3. Does Darboux’s Theorem require the derivative to be continuous?
No, Darboux’s Theorem does not require the derivative to be continuous. The only requirement is that the function f is differentiable on the interval.
- A derivative may have discontinuities.
- However, it cannot have jump discontinuities.
- This is because it must still take all intermediate values.
4. What is the difference between Darboux’s Theorem and the Intermediate Value Theorem?
The key difference is that the Intermediate Value Theorem (IVT) applies to continuous functions, while Darboux’s Theorem applies specifically to derivatives.
- IVT: If a function is continuous on [a, b], it takes every value between f(a) and f(b).
- Darboux’s Theorem: If f is differentiable, then f′ takes every value between f′(a) and f′(b).
5. Can you give an example illustrating Darboux’s Theorem?
An example of Darboux’s Theorem is when a derivative changes from negative to positive values, it must pass through zero. Suppose:
- f′(a) = -2
- f′(b) = 3
6. Why is Darboux’s Theorem important in real analysis?
Darboux’s Theorem is important because it shows that derivatives cannot have jump discontinuities. Its significance includes:
- Helping prove the existence of critical points.
- Supporting results in mean value theorems.
- Clarifying the structure of differentiable functions.
7. What are the conditions required for Darboux’s Theorem?
The main condition for Darboux’s Theorem is that the function must be differentiable on an open interval (a, b). Specifically:
- f must be differentiable on (a, b).
- No assumption of continuity of f′ is required.
8. Can a derivative be discontinuous and still satisfy Darboux’s Theorem?
Yes, a derivative can be discontinuous and still satisfy Darboux’s Theorem because it must still have the Intermediate Value Property.
- Derivatives may have removable or oscillatory discontinuities.
- They cannot have jump discontinuities.
- This is because they cannot skip intermediate values.
9. How is Darboux’s Theorem related to the Mean Value Theorem?
Darboux’s Theorem is closely related to the Mean Value Theorem (MVT) because both concern properties of derivatives on intervals.
- The MVT guarantees a point c where f′(c) = (f(b) − f(a))/(b − a).
- Darboux’s Theorem guarantees that derivatives take all intermediate values.
10. What is the Intermediate Value Property in Darboux’s Theorem?
The Intermediate Value Property means that a function takes every value between any two of its values on an interval. In Darboux’s Theorem:
- If f′(a) = m and f′(b) = n,
- Then for any k between m and n,
- There exists c in (a, b) such that f′(c) = k.
