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Mean Value Theorem
Theorem (Mean Value Theorem). Let $[ a, b ] \subset D(h)$, where $h$ is differentiable on $( a, b )$ and $h$ is continuous on $[ a, b ]$. Then
Corollary 1. Let $f: {\bf R} \rightarrow {\bf R}$ and $f'(x) = 0$, for all $x$ in an interval $I$ in ${\bf R}$.
Then $f$ is constant.
The condition implies that the derivative is zero, hence by the MVT, $f$ is constant.
Another theorem that follows from the IVT is this.
Corollary 2. Let $f′(x) = g′(x)$ for all $x$ in an open subset $U$ of ${\bf R}^n$ and
$$f(a) = g(a).$$
Then
In order to proof the Corollary, take $h = f - g$, then apply Corollary 1.
Same proof for vector functions.
Corollary 1'. Let $f: {\bf R}^n \rightarrow {\bf R}^m$
$$f'(x) = 0,$
for all $x$ in an open subset $U$ of ${\bf R}^n$.
Then $f$ is constant.
The condition implies that the gradient is zero, so all the partial derivatives $\frac{\partial f_j}{\partial x_i}$ are zero, hence by the MVT, every component function $h = f_j$ of $f$ is constant with respect to each of the variables. Hence it is constant and $f$ is constant too.
Corollary 2'. Let $f'(x) = g'(x)$ for all $x$ in an open subset $U$ of ${\bf R}^n$ and
$$f(a) = g(a).$$
Then
The Mean Value Theorem*
The Fermat's Theorem is an example of a theorem the converse of which isn't true. This is what it does and does not say about a function differentiable at $X=C$: $$X=C \text{ is }\ \begin{array}{lll} \text{a local}&\Longrightarrow\\ \text{max/min }&\not\Longleftarrow \end{array} \quad \nabla f\, (C)=0.$$
Let's interpret the conclusion in terms of terrains:
- $X$ is location on the map,
- $f(X)$ is the elevation at location $X$,
- $\nabla f\, (X)$ is the slope of the terrain at location $X$.
Then we conclude that
- $(\Rightarrow )$ whenever we are the highest from the sea level, there is no sloping.
But not conversely:
- $(\not\Leftarrow )$ even if there is no sloping at the location, we can resume climbing in the same direction.
Let's make this observation mathematical and turn it into a theorem. We again assume that $X$ is time, limited to interval $[a,b]$.
Theorem (Rolle's Theorem). Suppose $R$ is a closed bounded region in ${\bf R}^n$ and suppose function $z=f(X)$ satisfies the following:
- 1. $f$ is continuous on $R$,
- 2. $f$ is differentiable inside $R$, which is assumed to be non-empty,
- 3. $f(X) = 0$ for all $X$ in $\partial R$.
Then $\nabla f\, (C) = 0$ for some $C$ inside $R$.
In the theorem, #1 means that the mountain has no cliffs, #2 means that it has no sharp ridges or peaks, #3 means that it starts at the sea level.
Proof. The proof repeats the proof of the original Rolle's Theorem from Part I. Suppose $f$ has on $R$:
- a global maximum at $X = C$ and
- a global minimum at $X = D$.
These assumptions are justified by the Extreme Value Theorem (that's why we need to assume continuity).
Case 1: Either $C$ or $D$ is not on the boundary of $R$. We now use the Global Extrema Theorem: every global extreme point has $0$ gradient when it's not on the boundary of the region. It follows that $\nabla f\, (C) = 0$ or $\nabla f\, (D) = 0$, and we are done.
Case 2: Both $C$ and $D$ are on the boundary of $R$. Then $f(C) = 0$ and $f(D) = 0$. They are equal, which means that: $$\max f= \min f !$$ Therefore $f$ is constant on $R$, so $\nabla f\, (X) = 0$ for all $X$ inside $R$. $\blacksquare$
The condition of the theorem simply states that the difference quotient is zero, $$\frac{\Delta f}{\Delta X}=0,$$ where $R$ is a square and the partition of $R$ is trivial: $n=1$.
Thus, Rolle's Theorem says that if the mountain stands on a horizontal plane, it has horizontal locations. In other words, $f(X)$ is constant on the boundary of $R$. Another interpretation of this is: the change of the elevation along any trip across the mountain is zero.
Thus, Rolle's Theorem says:
- the average rate of change from boundary to boundary is zero $\Longrightarrow$ the instantaneous rate of change is zero at some point; or
- the slopes of all secant lines from boundary to boundary are zero $\Longrightarrow$ the slope of the tangent plane is zero at some point.
Now, do the hypothesis and the conclusion have to be with respect to a horizontal plane? Can it be inclined?
What happens if we rotate (or skew) the graph?
The picture suggests what happens to the entities we considered in Rolle's theorem:
- the tangents at those special points that used to be horizontal have become inclined;
- the secant line that connects the end-points that used to be horizontal has become inclined.
But these lines are still parallel!
Theorem (Mean Value Theorem). Suppose $R$ is a closed bounded region in ${\bf R}^n$ and suppose function $z=f(X)$ satisfies the following:
- 1. $f$ is continuous on $R$,
- 2. $f$ is differentiable inside $R$.
Suppose for some linear function $z=l(X)$, we have $$f(X)=l(X), \text{ for all } X \text{ in } \partial R.$$ Then $$\nabla l = \nabla f\, (C),$$ for some $C$ inside $R$.
Note that MVT is more general than RT. In other words, the latter is an instance, a narrow case of MVT. The proof of MVT however will rely on RT. We just “skew” the graph of MVT back to RT:
Proof. The proof repeats the proof of the original Mean Value Theorem from Part I. Let's rename $f$ in Rolle's Theorem as $h$ to use it later. Then its conditions take this form:
- 1. $h$ continuous on $R$.
- 2. $h$ is differentiable inside $R$.
- 3. $h(X)=0$ on $\partial R$.
Now back to $f$. The key step is to define a new function: $$h(X) = f(X) – l(X),$$ and verify the conditions above. 1. $h$ is continuous on $R$ as the difference of the two continuous functions (SR). 2. $h$ is differentiable inside $R$ as the difference of the two differentiable functions (SR) and: $$\nabla h\, (X)=\nabla f\, (X)-\nabla l.$$ 3. We have on $\partial R$: $$f(X) = l(X)\ \Longrightarrow\ h(X) = 0.$$ Thus, $h$ satisfies the conditions of RT. Therefore, the conclusion is satisfied too: $$\nabla h\, (C)=0$$ for some $C$ in $R$. In other words, we have $$\nabla f\, (C)-\nabla l=0.$$ $\blacksquare$
Geometrically, $C$ is found by shifting the secant plane until it touches the graph: