Derivative Function
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's Theorem
1. What Does the Darboux Theorem States?
Darboux theorem states that if f is differentiable on closed interval [x,y], and r is any number between f’(x) and f’(y), then there includes a number x in the open interval (x,y) such that f’(z) = r
2. What is an Intermediate Value Theorem?
The intermediate theorem states that if a continuous function attains two values, then it must attain all the values in between these values. A continuous function is a function whose graph can be drawn without pulling a pencil from the paper. For examples, if f(x) is continuous functions that join the point [0,0] and [6,20], then there must be xm between 0 and 6 where y = 1 and also some xn where y = 19 where y = 19/6, etc. For this function, there is an x ∈ [0, 6] for any value between 6 and 20.
3. What is Intermediate Value Property or Darboux Property?
A function f : A → K is said to have darboux property or intermediate value property, on a set B ⊆ A, Iff together with any two functional values f(p) and f(p)1, (p, p1 ∈ B), it also consider all the intermediate values between f(p) and f(p)1 at some points of B.
In other words, the image set of f[B] includes the entire interval between f(p) and f(p)1 in E.
4. Who Introduced Darboux Theorem?
A French Mathematician Jean - Gaston Darbouzx first proved the Darboux theorem in the 19th century
