Is there a point where the value of the function $g$ is greater than the length of the interval? Prove or disprove.
Suppose that $g$ is a positive real valued function of a real number. If $a < b$ are real numbers, then there is a finite sequence $a = t_0 < t_1 \dots < t_n = b$ of real numbers such that in each interval $[t_k,t_{k+1}]$ there is a point where the value of the function $g$ is greater than the length of the interval. 
Unfortunately, I am quite lost. I tried a contradiction proof, but I ran into a snag. Any help would be welcome.
 A: It's a compactness argument.
For each $x$, let $B(x)$ be the open ball centered at $x$ of radius $\frac12f(x)$. Since $[a,b]$ is compact, it can be covered by some finite set $B(x_i)$ of these balls. Prune the list, removing any ball entirely contained in another. Now, put them in order so that $x_1<x_2<x_3<\dots<x_{N-1}$.
In this form, we claim that (1) $a\in B(x_1)$, (2) $B(x_i)$ and $B(x_{i+1})$ intersect for $i=1,2,\dots,N-2$, and (3) $b\in B(x_{N-1})$. Why? In order for the balls to not contain each other, the lower endpoints $x_i-\frac12f(x_i)$ and upper endpoints $x_i+\frac12f(x_i)$ must each be in the same order as the $x_i$. Thus $x_1-\frac12f(x_1)$ is the smallest lower endpoint - and since $a$ is in the union of the balls, $a>x_1-\frac12f(x_1)$ is in $B(x_1)$, proving (1). (3) is similar. For (2), $\cup_{j\le i}B(x_j)\subseteq (x_1-\frac12f(x_1),x_i+\frac12f(x_i))$ and $\cup_{j>i}B(x_j) \subseteq (x_{i+1}-\frac12f(x_{i+1}),x_{N-1}+\frac12f(x_{N-1}))$. Since the $B(x_j)$ cover the interval, the endpoints must cross; $x_i+\frac12f(x_i)>x_{i+1}-\frac12f(x_{i+1})$, making $B(x_i)$ and $B(x_{i+1})$ intersect.
Now, we're almost there. Let $t_0=a,t_1=x_1,t_2=x_2,\dots,t_{N-1}=x_{N-1},t_N=b$.
On the interval $[t_0,t_1]$, $f$ takes the value $f(x_1)$ at the upper endpoint. Since $a\in B(x_1)$, $t_1-t_0=x_1-a < \frac12f(x_1)$. Check.
On the interval $[t_{N-1},t_N]$, $f$ takes the value $f(x_{N-1})$ at the lower endpoint. Since $b\in B(x_{N-1})$, $t_N-t_{N-1}=b-x_{N-1} < \frac12f(x_{N-1})$. Check.
On a middle interval $[t_i,t_{i+1}]$, $f$ takes values $f(x_i)$ and $f(x_{i+1})$ at the endpoints. Since $B(x_i)$ and $B(x_{i+1})$ overlap, $t_{i+1}-t_i = x_{i+1}-x_i < \frac12f(x_i)+\frac12f(x_{i+1})\le \max(f(x_i),f(x_{i+1}))$. Check.
For every subinterval of this partition, we have verified that the length of the subinterval is less than the value of $f$ at one of the subinterval's endpoints.
