Continuous Calculus

The condition that $(x - \frac{1}{2} ) ( f(x) - \frac{x}{2} ) > 0$ means that $x - \frac{1}{2}$ and $f(x) - \frac{x}{2}$ must have the same sign.
If $x < \frac{1}{2}$ , then $f(x) < \frac{x}{2}$ .
If $x > \frac{1}{2}$ , then $f(x) > \frac{x}{2}$ .
We summarize this pictorially by seeing that $f(x)$ must lie in the shaded region on the right.

At $x = \frac{1}{2}$ we see that the graph has to be "somewhat steep", as opposed to "somewhat flat". This causes us to believe that if $f' ( \frac{1}{2} )$ exists, then the tangent at $x = \frac{1}{2}$ must lie within the shaded region, hence the slope cannot be smaller than $\frac{1}{2}$ . Put another way, the intuitive reason is that if the slope is flatter than $\frac{1}{2}$ , then there is a point $x' > \frac{1}{2}$ such that $\frac{ f(x') - f( \frac{1}{2} ) } { x' - \frac{1}{2} } < \frac{1}{2}$ . Let's rigorize this claim.

First, we will show that $f( \frac{1}{2} ) = \frac{1}{4}$ .
Proof by contradiction. Since $f$ is a continuous function, so $f( \frac{1}{2} )$ exists. Suppose it is not equal to $\frac{1}{4}$ , then without loss of generality, $f( \frac{1}{2} ) = Y > \frac{1}{4}$ .
For $x < \frac{1}{2}$ , we have $2x - 1 < 0$ and so $2 f(x) - x < 0 \Rightarrow f(x) < \frac{x}{2} < \frac{1}{4}$ . As such, this implies that the left hand limit:
$\lim_{ x \rightarrow \frac{1}{2} ^ - } f(x) \leq \frac{1}{4} < Y = f ( \frac{1}{2} )$ which contradicts the fact that the limit of the continuous function is $f(\frac{1}{2} )$ .

Second, we will show that $f'( \frac{1}{2} ) \geq \frac{1}{2}$ .
Proof by contradiction. Suppose that $f' ( \frac{1}{2} )$ exists and (without loss of generality) is equal to $S < \frac{1}{2}$ . Then, by the epsilon-delta definition of a limit , we know that for $\epsilon = \frac{ \frac{1}{2} - S } { 2 }$ (see note below for why we made this choice of $\epsilon$ ), there exists a $\delta$ such that for any $x' \in ( \frac{1}{2} - \delta, \frac{1}{2} + \delta)$ , we have

$\left|\frac { f(x') - \frac{1}{4} } { x' - \frac{1}{2} } - S \right| < \epsilon \Rightarrow S - \epsilon < \frac { f(x') - \frac{1}{4} } { x' - \frac{1}{2} } < S + \epsilon < \frac{1}{2}$

(Note: At this point, we have the mathematical statement of our "intuitive reason". For rigor, we need to link it back to the conditions in the problem, instead of just our graphical image.)
Fix an $x' \in ( \frac{1}{2} , \frac{1}{2} + \delta )$ , so $(2x' - 1) > 0$ . From the above, we get that $f(x') < \frac{1}{2} ( x' - \frac{1}{2}) + \frac{1}{4} ) = \frac{1}{2} x'$ , thus $(2 f(x') -x') < 0$ . This contradicts $( 2x'-1) ( 2f(x') -x') > 0$ .

Note: Reviewing the second proof, the main/only usage of $\epsilon$ is in requiring that $S + \epsilon < \frac{1}{2}$ . Any $\epsilon < \frac{1}{2} - S$ will work, and I chose $\epsilon = \frac{ \frac{1}{2} - S } { 2 }$ .

TRUE: $f'(\frac{1}{2}) \ge \frac{1}{2}$ . To prove this, we will need the following lemma:

Lemma: $f(\frac{1}{2}) = \frac{1}{4}$ .

Proof of lemma: Let $g(x) = (2x - 1)(2f(x) - x)$ . By hypothesis $g(x) > 0$ for $x \neq \frac{1}{2}$ , and $g(\frac{1}{2}) = 0$ . Also by hypothesis $f$ is continuous on $[0, 1]$ and differentiable at $\frac{1}{2}$ , so $g$ also has these properties. Thus $g$ has a local minimum at $x = \frac{1}{2}$ , where it is differentiable, which means $g'(\frac{1}{2}) = 0$ . Since

$g'(x) = (2x - 1)(2f'(x) - 1) + 2(2f(x) - x)$ ,

we have

$0 = g'(\frac{1}{2}) = 4f(\frac{1}{2}) - 1$ ,

hence $f(\frac{1}{2}) = \frac{1}{4}$ .

Alternate proof of lemma: Because $(2x - 1)(2f(x) - x) > 0$ for $x \neq \frac{1}{2}$ , $2x-1$ and $2f(x) - x$ have the same sign, which means $f(x) < \frac{x}{2}$ for $0 \le x < \frac{1}{2}$ and $f(x) > \frac{x}{2}$ for $\frac{1}{2} < x \le 1$ . Therefore by the continuity of $f(x)$ and $\frac{x}{2}$ ,

$\frac{1}{4} = \lim_{x \to \frac{1}{2}^+} \frac{x}{2} \le \lim_{x \to \frac{1}{2}^+} f(x) = \lim_{x \to \frac{1}{2}} f(x) = \lim_{x \to \frac{1}{2}^-} f(x) \le \lim_{x \to \frac{1}{2}^-} \frac{x}{2} = \frac{1}{4}$ .

Thus by the squeeze theorem, $f(\frac{1}{2}) = \lim_{x \to \frac{1}{2}} f(x) = \frac{1}{4}$ .

Proof of solution: As noted above, $2x-1$ and $2f(x) - x$ have the same sign, so $\frac{2f(x) - x}{2x - 1} > 0$ for $x \neq \frac{1}{2}$ . Hence:

$\begin{aligned} f'(\frac{1}{2}) &= \lim_{x \to \frac{1}{2}} \frac{f(x) - f(1/2)}{x - 1/2}\\ &= \lim_{x \to \frac{1}{2}} \frac{2f(x) - 2f(1/2)}{2x - 1}\\ &= \lim_{x \to \frac{1}{2}} \frac{2f(x) - 1/2}{2x - 1}\\ &= \lim_{x \to \frac{1}{2}} \frac{2f(x) - x + x - 1/2}{2x - 1}\\ &= \lim_{x \to \frac{1}{2}} \frac{2f(x) - x}{2x - 1} + \lim_{x \to 1/2} \frac{x - 1/2}{2x - 1}\\ &\ge \lim_{x \to \frac{1}{2}} \frac{x - 1/2}{2x - 1} = 1/2 \quad\square\\ \end{aligned}$