- #1

- 940

- 59

- 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.

(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.