ANA1_L24

Mean value theorems

Derivative is zero at local extrema

Rudin, 5.7

Definition 1(Definition).

Let $f$ be a real function defined on a metric space $X$. We say that $f$ has a local maximum at a point $p\in X$ if there exists a $\delta >0$ such that for all $x\in B_{\delta }(p)$ we have$f(x)\le f(p)$.
Local minima are defined similarly.

Rudin, 5.8

Theorem 2(Theorem).

Let $f:[a,b]\to \mathbb{R}$, $p\in (a,b)$. If $f$ has a local maximum at $p$ and $f^{\prime }(p)$ exists, then $f^{\prime }(p)=0$.

Proof
Consider $\delta$ such that $B_{\delta }(p)\subset (a,b)$.
$t\in (p-\delta ,p)$ $⟹$ $\frac{f(t)-f(p)}{t-p}\ge 0$ $⟹$ $f^{\prime }(p)\ge 0$.
$t\in (p,p+\delta )$ $⟹$ $\frac{f(t)-f(p)}{t-p}\le 0$ $⟹$ $f^{\prime }(p)\le 0$.
Thus, $f^{\prime }(p)=0$. ❏

Rolle’s theorem

Theorem 3(Theorem).

Suppose $f:[a,b]\to \mathbb{R}$ is continuous, $f$ is differentiable on $(a,b)$, and $f(a)=f(b)$. Then, there exists $p\in (a,b)$ such that $f^{\prime }(p)=0$.

Proof
We know from the extreme value theorem that $f$ must attain a maximum and a minimum value. If $f$ is constant, any $p\in (a,b)$ works. Otherwise, there exists $x\in (a,b)$ such that $f(x)\ne f(a)=f(b)$. If $f(x)>f(a)$ then take $p\in (a,b)$ such that $f(p)$ is the global maximum, else take $p\in (a,b)$ such that $f(p)$ is the global minimum. From 5.8, $f^{\prime }(p)=0$. ❏

Mean value theorem

Rudin, 5.10

Theorem 4(Theorem).

Suppose $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$.
Then, there exists $p\in (a,b)$ such that

$$f(b)-f(a)=(b-a)f^{\prime }(p).$$

Proof
Rolle’s theorem can be made applicable here by subtracting the line joining $(a,f(a))$ and $(b,f(b))$ from $f(x)$, which makes the value at $a$ and $b$ equal.

$$\begin{array}{rl}h(x)& \equiv f(x)-\left[f(a)+\left(\frac{f(b)-f(a)}{b-a}\right)(x-a)\right]\\ & =\frac{1}{b-a}\left[(b-a)(f(x)-f(a))-(f(b)-f(a))(x-a)\right]\end{array}$$

Note that $h(a)=h(b)=0$. From Rolle’s theorem, there exists $p\in (a,b)$ such that $h^{\prime }(p)=0$.

$$\begin{array}{rl}& h^{\prime }(x)=\frac{1}{b-a}\left[(b-a)f^{\prime }(x)-(f(b)-f(a))\right]\\ & \\ & h^{\prime }(p)=0⟹f(b)-f(a)=(b-a)f^{\prime }(p)\end{array}$$

❏

Generalized mean value theorem

Rudin, 5.9

Theorem 5(Theorem).

If $f,g:[a,b]\to \mathbb{R}$ are continuous on $[a,b]$ and differentiable on $(a,b)$, then there is a point $p\in (a,b)$ such that

$$[f(b)-f(a)]g^{\prime }(p)=[g(b)-g(a)]f^{\prime }(p).$$

Notice that plugging $g(x)=x$ gives us the mean value theorem.

Proof
If $f(a)=f(b)$, the proof follows trivially from Rolle’s theorem. Ditto if $g(a)=g(b)$. Thus, assume $f(a)\ne f(b)$ and $g(a)\ne g(b)$. The strategy is similar to that of the previous proof:

$$\begin{array}{rl}h(x)& \equiv f(x)-\left[f(a)+\left(\frac{f(b)-f(a)}{g(b)-g(a)}\right)(g(x)-g(a))\right]\\ & =\frac{1}{g(b)-g(a)}\left[(g(b)-g(a))(f(x)-f(a))-(f(b)-f(a))(g(x)-g(a))\right]\end{array}$$

Note that $h(a)=h(b)=0$. From Rolle’s theorem, there exists $p\in (a,b)$ such that $h^{\prime }(p)=0$.

$$\begin{array}{rl}& h^{\prime }(x)=\frac{1}{g(b)-g(a)}\left[(g(b)-g(a))f^{\prime }(x)-(f(b)-f(a))g^{\prime }(x)\right]\\ & h^{\prime }(p)=0⟹(g(b)-g(a))f^{\prime }(p)=(f(b)-f(a))g^{\prime }(p).\end{array}$$

❏

Remark

A simpler version of $h(x)$ that can be used instead:

$$h(x)\equiv (f(b)-f(a))g(x)-(g(b)-g(a))f(x)$$

Rudin, 5.11

Rudin, 5.11

Theorem 6(Theorem).

Suppose $f$ is differentiable in $(a,b)$. Then,

1. If $f^{\prime }(x)\ge 0$ for all $x\in (a,b)$, then $f$ is monotone increasing.
2. If $f^{\prime }(x)=0$ for all $x\in (a,b)$ then $f$ is constant.
3. If $f^{\prime }(x)\le 0$ for all $x\in (a,b)$, then $f$ is monotone decreasing.

Proof
For (1), Let $x_2,x_1\in (a,b)$, $x_1<x_2$. Then, $f(x_2)-f(x_1)=(x_2-x_1)f^{\prime }(x)$ for some $x\in (x_1,x_2)$. Thus, $f(x_2)-f(x_1)\ge 0$. Similar proofs for (2) and (3). ❏

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Darboux’s Theorem

Not all functions can be derivatives. While every derivative certainly need not be continuous (for example, $x^2\sin \left(\frac{1}{x}\right)$ is differentiable at every point in $\mathbb{R}$, but its derivative is not continuous at $0$), every derivative shares a property with continuous functions, namely, the intermediate value property.

Rudin, 5.12

Theorem 7(Theorem).

Suppose $f$ is a real differentiable function on $[a,b]$ and suppose WLOG $f^{\prime }(a)<\lambda <f^{\prime }(b)$. Then there is a point $x\in (a,b)$ such that $f^{\prime }(x)=\lambda$.

Proof
Define $g(t)\equiv f(t)-\lambda t$. Notice that $g^{\prime }(a)<0$ and $g^{\prime }(b)>0$. This implies that there exist $t_1,t_2\in (a,b)$ such that $g(t_1)<g(a)$ and $g(t_2)<g(b)$. Thus, $g$ attains its minimum value at some $x\in (a,b)$. Since the derivative is zero at local extrema, we have $g^{\prime }(x)=0$, which gives us $f^{\prime }(x)=\lambda$. ❏

Info

How does $g^{\prime }(a)<0$ imply the existence of $t_1\in (a,b)$ such that $g(t_1)<g(a)$? Recall the definition of limit of a function. We know that ${\lim }_{x\to a}\frac{g(x)-g(a)}{x-a}<0$. Choose $ϵ$ to be $|g^{\prime }(a)|$, and obtain the corresponding $\delta$. So, for all $x\in (a,a+\delta )$, $\frac{g(x)-g(a)}{x-a}\in (-2g^{\prime }(a),0)$, i.e, $g(x)<g(a)$. Note that we arrived at stronger statement than we required: $g^{\prime }(a)<0$ not only implies the existence of $t_1\in (a,b)$, but also the existence of an entire interval $(a,a+\delta )$ satisfying the same condition.

Theorem 8(Corollary).

If $f$ is differentiable on $[a,b]$, then $f^{\prime }$ cannot have any discontinuities of the first kind on $[a,b]$.

Proof
We will prove more generally that any real function $f$ that satisfies the intermediate value property on $[a,b]$ cannot have discontinuities of the first kind. FTSOC, assume $f$ has such a discontinuity at $p\in [a,b]$. Let the left hand limit at $p$ be $L$, and the right hand limit at $p$ be $R$. Either $L\ne R$, or $L=R$ but $f(p)\ne L$.
In the first case, let $d=|L-R|$. Let $ϵ=d/2$. Find ${\delta }_1$ and ${\delta }_2$ such that $x\in (p-{\delta }_1,p)$ $⟹$ $f(x)\in (L-ϵ,L+ϵ)$ and $x\in (p,p+{\delta }_2)$ $⟹$ $f(x)\in (R-ϵ,R+ϵ)$. Pick $t_1\in (p-{\delta }_1,p)$ and $t_2\in (p,p+{\delta }_2)$. Now, $f$ satisfies the intermediate value property on $[t_1,t_2]$. WLOG, $f(t_1)<\frac{L+R}{2}<f(t_2)$. But, $\nexists$ $t\in (t_1,t_2)$ such that $f(t)=\frac{L+R}{2}$ (If it happens that $f(p)=\frac{L+R}{2}$, one can easily find another such point that $f(p)$ now cannot be equal to). $\Rightarrow \Leftarrow$
In the second case, let $d=|L-f(p)|$. Let $ϵ=d/2$. Find $\delta$ such that $x\in (p-\delta ,p+\delta )$ $⟹$ $f(x)\in (L-ϵ,L+ϵ)$. Choose $t_1\in (p-\delta ,p)$. $f$ satisfies the intermediate value property on $[t_1,p]$. WLOG, $f(t_1)<L+ϵ<f(p)$. But, $\nexists$ $t\in (t_1,p)$ such that $f(t)=L=ϵ$. $\Rightarrow \Leftarrow$ ❏

$f^{\prime }$ may have discontinuities of the second kind, though.

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Epilogue

Find a differentiable function $f$ and a point $p$ such that $f^{\prime }(p)>0$ but the function is not increasing in any interval containing $p$.

Consider

$$f(x)=\{\begin{array}{ll}x+2x^2\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0& x=0.\end{array}$$

It can be shown that $f$ is differentiable on $\mathbb{R}$ and $f^{\prime }(0)=1$. The derivative when $x\ne 0$ is

$$f^{\prime }(x)=1+4x\sin \left(\frac{1}{x}\right)-2\cos \left(\frac{1}{x}\right).$$

Now, we know that

$$\underset{x\to 0}{\lim }4x\sin \left(\frac{1}{x}\right)=0.$$

Consider any interval $[t_1,t_2]$ containing $0$. Let $ϵ=1$. Find $\delta >0$ such that $(-\delta ,\delta )\subset [t_1,t_2]$ and $x\in (-\delta ,\delta )⟹4x\sin \left(\frac{1}{x}\right)\in (-1,1)$. Find an integer $n$ such that

$$\frac{1}{2n\pi }\in (-\delta ,\delta ).$$

Thus, we have

$$f^{\prime }\left(\frac{1}{2n\pi }\right)<1+1-2=0.$$

Thus, $f$ is not increasing on $[t_1,t_2]$.