Proving the Existence and Value of a Derivative at a Point

[Eclair_de_XII]

TL;DR Summary

Let $f$ be a differentiable function on the interval $[a,b]$ and with the property that $f'(a)>0>f'(b)$. Prove that there is a point $c\in (a,b)$ with the property that $f'(c)=0$. Lemmas invoked: $f'>0$ means increasing, sign of terms of sequence converging to some positive number are also positive.

This proof has three steps and is very similar to (if not the same as) that other proof I posted here.(1) Prove the existence of a ball centered around $a$ with the property that $f'$ evaluated at any point in the ball is positive.

(2) Prove that the right end-point of this ball is bounded from above.

(3) Determine the value of the derivative evaluated at the supremum of the right end-points of the ball.===1===

Suppose this were not the case. Set $\epsilon=f'(a)$ and let $\delta>0$. It follows that for every $x\in(a,a+\delta)$, $f'(x)\leq 0$.Denote $T(x)=\frac{f(x)-f(a)}{x-a}\leq 0$. Hence, $x-a>0$ and $f(x)-f(a)\leq 0$, which means that $T(x)\leq 0$.\begin{align*}

|f'(a)-T(x)|&=&f'(a)+(-T(x))\\

&=&f'(a)+T(x)\\

&\geq&f'(a)\\

&=&\epsilon

\end{align*}Hence, there is a $\delta>0$ such that for $x\in (a,a+\delta)$, $f'(x)>0$.Similar reasoning shows that there is a ball centered around $b$ s.t. $f'$ evaluated at any point in this ball is negative.===2===

If $\delta=b-a$, then for $x\in(a,a+\delta)=(a,b)$, $f'(x)>0$. This contradicts the fact that there is a ball around $b$ whose points evaluate to negative values on $f'$. Let $t=\sup \{\delta>0:(a,a+\delta)\}$===3===

Let $x_n$ be a sequence in $(a,a+\delta)$ converging to $t$. Denote $S(n)=\frac{f(t)-f(x_n)}{t-x_n}$. Note that $f(t)-f(x_n)>0$ and $t-x_n>0$. Hence, $S(n)>0$. This is a sequence of positive numbers, which means that $S(n)$ could not converge to a negative number, which is to say, that $f'(t)\geq 0$.Now consider a sequence $y_n\in(a+t,b)$. Denote $P(n)=\frac{f(t)-f(y_n)}{t-y_n}$. Note that $f'(y)\leq0$ for $y\in(a+t,b)$. Hence, $0>t-y_n$ and $f(t)\geq f(y_n)$, and so $P(n)\geq0$. It follows that the sequence $P(n)$ could not converge to a negative number, which is to say that $f'(t)\leq 0$.$f'(t)=0$ as a result.

 

[Infrared]

(1) Prove the existence of a ball centered around $a$ with the property that $f'$ evaluated at any point in the ball is positive.

This claim is false. A counterexample is $f(x)=x^2\sin(1/x)+x/100$ for $x\neq 0$ with $f(0)=0.$ You can check that $f'(0)=1/100$, but $f'(x)$ takes on both positive and negative values in any neighborhood of $0.$

The first problem with your attempt proof is that the negation of "$f'(t)>0$ for all $t$ in a neighborhood of $a$" is not "$f'(t)\leq 0$ for all $t$ in a neighborhood of $a$ with $t\neq a$" but instead "In any neighborhood of $a$, there exists a $t$ such that $f'(t)\leq 0$."

Still, you're on the right track. Think about where the maximum of $f$ can occur (there must be one since $f$ is differentiable and hence continuous). Show that it can't occur at either endpoint, and hence has to occur at an interior point, where the derivative must be zero.

 

[Svein]

To the OP: In your proof you imply that f is is a continuously differentiable function on the interval [a,b]. This is a much stricter requirement than that f is differentiable.

 

[pasmith]

If $f(a) = f(b)$ then you can apply Rolle's theorem. Otherwise it seems that the conditions on $f'(a)$ and $f'(b)$ mean that you can find a subinterval of $[a,b]$ on which you can apply Rolle's theorem.

 

[Eclair_de_XII]

Lemma. There is a point $x>a$ s.t. $f(x)>f(a)$.

Proof. Since $f$ is differentiable, $\forall\epsilon>0$, $\exists\delta>0$ s.t. if $|h|<\delta$, $|\frac{f(a+h)-f(a)}{h}-f'(a)|<\epsilon$. Set $x=a+h$ where $h\in(0,\delta)$. It follows then that:

\begin{align*}
|[f(a)+f'(a)h]-f(a+h)|&=&\left|f(a)+\left[\frac{f(a+h)-f(a)}{h}\right]h-f(a+h)\right|\\
&=&|f(a)+[f(a+h)-f(a)]-f(a+h)|\\
&=&0\\
&<&\epsilon
\end{align*}

Hence, $f(a+h)\rightarrow f(a)+f'(a)h>f(a)$.

===

Theorem. If $f$ is differentiable on the interval $[a,b]$ and $f'(a)>0>f'(b)$, then there is a point $c\in(a,b)$ s.t. $f'(c)=0$.

Proof. By the lemma, there is a point $x_1>$ s.t. $f(x_1)>f(a)$. If $f'(x_1)>0$, then there is a point $x_2>x_1$ s.t. $f(x_2)>f(x_1)$. Otherwise, there is a point $x_2<x_1$ $f(x_2)\geq f(x_1)$. By construction, we have a sequence of points $\{x_n\}$ and an increasing sequence of points $\{f(x_n)\}$.

Note: If $f'(x_i)>0$ for all natural numbers $i$, then $\{f'(x_n)\}$ is a sequence of positive numbers converging to $f'(b)<0$, which is a contradiction. Hence, it is always possible to find a subsequence of $\{f'(x_n)\}$ whose terms are non-positive.

Note: Invoking the fact that the closed interval $[a,b]$ is sequentially compact.

There is a subsequence of $\{x_n\}$, call it $\{x_{n_k}\}$, that converges to some point $x\in(a,b)$. Since $f$ is continuous, we have that $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ s.t. if $n_k\geq N$, then $|f(x_{n_k})-f(x)|<\epsilon$.

As a result, for any $\epsilon>0$, $f(x)>f(x_{n_k})-\epsilon$, which means that $\epsilon>f(x_{n_k})-f(x)$. In turn, no term of the subsequence $\{f(x_{n_k})\}$ must exceed $f(x)$. Hence, $f(x)\geq f(x_{n_k})$ for all $n_k\geq N$.

Since $\{x_n\}\subset (a,b)$, it follows that the sequence, and by proxy, any of its subsequences, is bounded. Let $s\equiv\inf\{x_{n_k}\}_{n_k\geq N}$ and $t\equiv\sup\{x_{n_k}\}_{n_k\geq N}$. Hence, $f(x)\geq f(y)$ for $y\in(s,t)$. Note also that if $x\notin (s,t)$, the point $x$ would fail to lie within the interval of convergence.

 

[Office_Shredder]

I think there are multiple issues. For starters,

Hence, $f(a+h)\rightarrow f(a)+f'(a)h>f(a)$.

This is kind of sloppy. You should be able to prove very precisely and concisely that $f(a+h)>f(a)$. Given your other proof attempts, I think you would benefit from the exercise. The basic idea is since the limit of the derivative definition is positive, there must be small $h>0$ such that $(f(a+h)-f(a))/h > f'(a)/2 >0$. Then since $h>0$, $f(a+h)>f(a)$. You might want to clean up the details here.

.

Proof. By the lemma, there is a point $x_1>$ s.t. $f(x_1)>f(a)$. If $f'(x_1)>0$, then there is a point $x_2>x_1$ s.t. $f(x_2)>f(x_1)$. Otherwise, there is a point $x_2<x_1$ $f(x_2)\geq f(x_1)$. By construction, we have a sequence of points $\{x_n\}$ and an increasing sequence of points $\{f(x_n)\}$.

This all looks true so far.

Note: If $f'(x_i)>0$ for all natural numbers $i$, then $\{f'(x_n)\}$ is a sequence of positive numbers converging to $f'(b)<0$, which is a contradiction. Hence, it is always possible to find a subsequence of $\{f'(x_n)\}$ whose terms are non-positive.

You are assuming f' is continuous to get that contradiction, which it might not be. Also, you are assuming $x_i$ converges to $b$ I think, which it might not.

Note: Invoking the fact that the closed interval $[a,b]$ is sequentially compact.

There is a subsequence of $\{x_n\}$, call it $\{x_{n_k}\}$, that converges to some point $x\in(a,b)$. Since $f$ is continuous, we have that $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ s.t. if $n_k\geq N$, then $|f(x_{n_k})-f(x)|<\epsilon$.

As a result, for any $\epsilon>0$, $f(x)>f(x_{n_k})-\epsilon$, which means that $\epsilon>f(x_{n_k})-f(x)$. In turn, no term of the subsequence $\{f(x_{n_k})\}$ must not exceed $f(x)$. Hence, $f(x)\geq f(x_{n_k})$ for all $n_k\geq N$.

There are a lot of negatives here that are confusing to me, but I don't see how you conclude this. It's equally true that $\epsilon > f(x)-f(x_{n_k})$

 

[Eclair_de_XII]

This is kind of sloppy. You should be able to prove very precisely and concisely that $f(a+h)>f(a)$. Given your other proof attempts, I think you would benefit from the exercise.

Would this work?

I wasn't sure if this proof made use of the fact that $f$ is continuously differentiable or just differentiable.

I think these are the last lines of the respective definitions?

(1) Continuously differentiable: ... $|f'(x)-f'(a)|<\epsilon$ for $x\neq a$
(2) Differentiable: ... $|\frac{f(x)-f(a)}{x-a}-f'(a)|<\epsilon$ for $x\neq a$

===

Let $\{a_n\}$ be a sequence of terms greater than $a$ and converging to $a$. Denote $T(a_n)\equiv \frac{f(a_n)-f(a)}{a_n-a}$.

By definition, for all $\epsilon>0$, there is an integer $N$ such that whenever $n\geq N$, $|T(a_n)-f'(a)|<\epsilon$, where $f'(a)>0$. If there is an integer $m\geq N$ s.t. $f(a_m)\leq f(a)$, then set $\epsilon=f'(a)$:

$a_m-a>0$
$f(a_m)-f(a)\leq 0$
$T(a_m)\leq 0$

\begin{align*}
|f'(a)-T(a_m)|&=&f'(a)+[-T(a_m)]\\
&=&f'(a)+|T(a_m)|\\
&\geq&\epsilon
\end{align*}

Hence, for all $n\geq N$, $f(a_n)>0$.

Choose $x=a_m$ for some $m\geq N$.

 

[Office_Shredder]

Byy definition, for all $\epsilon>0$, there is an integer $N$ such that whenever $n\geq N$, $|T(a_n)-f'(a)|<\epsilon$, where $f'(a)>0$. If there is an integer $m\geq N$ s.t. $f(a_m)\leq f(a)$, then set $\epsilon=f'(a)$

This is weirdly worded. Just say at the start, last $\epsilon = f'(a)$. You get to pick it, so just pick it.

$a_m-a>0$
$f(a_m)-f(a)\leq 0$
$T(a_m)\leq 0$

\begin{align*}
|f'(a)-T(a_m)|&=&f'(a)+[-T(a_m)]\\
&=&f'(a)+|T(a_m)|\\
&\geq&\epsilon
\end{align*}

Hence, for all $n\geq N$, $f(a_n)>0$.

Choose $x=a_m$ for some $m\geq N$.

I think you meant to say in the second to last line that $f(a_n)> f(a)$?

This looks right to me

 

[Eclair_de_XII]

I think you meant to say in the second to last line that $f(a_n)>f(a)$?

Yes.

 

[Eclair_de_XII]

===(continuing from the first paragraph of the second proof attempt in post #5)

Suppose that $f'(x_i)>0$ for all $i$. By construction, $\{x_n\}$ is an increasing sequence, and moreover, a sequence bounded from above by $b$. Denote $x\equiv\sup\{x_n\}$.

Note that the sequence of $x_n$ converges to $x$. (lemma)

Suppose $x>b$ and choose $d=x-b$. Since $\{x_n\}$ is a sequence in $(a,b)$, none of its terms exceed $b$. Hence, the set $B(x,d)\cap \{x_n\}=\varnothing$.

Now suppose that $x=b$. Then the sequence $\{x_n\}$ converges to $b$. Denote the difference quotient $T(x_n)\equiv\frac{f(b)-f(x_n)}{b-x_n}$ for $x_n\neq b$. Note:

$b-x_n>0$
$f(b)-f(x_n)>0$
$T(x_n)>0$

The sequence $\{T(x_n)\}$ converges to $f'(b)<0$, but every term of the sequence is positive. This is a contradiction as was shown earlier.

Hence, either $x<b$ or there are points $x_i$ s.t. $f'(x_i)\leq 0$.

If the former holds, then $f$ has a vertical asymptote at $x$, which means that it is not defined at $x$, so it cannot be continuous there.

If the latter holds, then by sequential compactness, there is a subsequence $\{x_{n_k}\}$ that converges to some point $x'$. Moreover, by continuity, $\{f(x_{n_k})\}$ is an increasing sequence that converges to $f(x')$; also, it is bounded for large enough $n_k$. Let $y\equiv \sup\{f(x_{n_k})\}$. If $y\neq f(x')$, then this contradicts the fact that bounded, increasing sequences converge to their supremum. Hence, $f(x')=y$.

Now consider $f'(x')$. If $f'(x')<0$, then consider the subsequence of $\{x_{n_k}\}$ consisting of terms less than $x'$.

$0<x'-x_{n_k}$
$f(x')-f(x_{n_k})>0$
$T(x_{n_k})>0$ but its limit is negative. This is a contradiction.

Same thing for terms of $\{x_{n_k}\}$ consisting of terms greater than $x'$.

 

[Svein]

Hm. Take a look at this curve: It is differentiable a. e. with derivatives jumping from +1 to -1 and back again. It is not hard to find an interval [a, b] with $f(a)<0 <f(b)$ but with no point c where $f'(c)=0$.

 

[Infrared]

@Svein Right, differentiable almost everywhere is not enough! The function really be differentiable on its whole domain. Similarly, the intermediate value theorem obviously fails for functions that are only continuous almost everywhere.