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Mean Value Theorem

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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

$\frac{h(b) - h(a)}{b - a} = h′(c)$ for some $c \in [ a, b ]$.

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

$f(x) = g(x)$ for all $x$ in $U$.

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

$f(x) = g(x)$ for all $x$ in $U$.

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.
Driving along y-axis.png

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.
Rolle's Theorem.png

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?

Rolle's Theorem rotated.png

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:

MVT proof.png

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:

MVT by shifting the secant.png