Let \(f:[0,1]\to[0,1]\) be \(C^1\), with \[ f(0)=f(1)=0, \] and suppose \(f'\) is nonincreasing, so that \(f\) is concave. Prove that the arclength of the graph of \(f\) does not exceed \(3\).
Solution
1. There exists \(c\in(0,1)\) such that \(f'(c)=0\).
Since \(f\) is continuous on \([0,1]\), differentiable on \((0,1)\), and \[ f(1)-f(0)=0, \] the mean value theorem gives \(c\in(0,1)\) with \[ f'(c)=\frac{f(1)-f(0)}{1-0}=0. \]
2. One has\[ f'(x)\ge0\quad(0\le x\le c), \qquad f'(x)\le0\quad(c\le x\le1). \]
The derivative \(f'\) is nonincreasing. Thus \(x\le c\) implies \[ f'(x)\ge f'(c)=0, \] while \(x\ge c\) implies \[ f'(x)\le f'(c)=0. \]
3. The total variation of \(f\) satisfies\[ \int_0^1\abs{f'(x)}\,dx=2f(c)\le2. \]
By step 2 and the fundamental theorem of calculus, \[ \begin{aligned} \int_0^1\abs{f'(x)}\,dx &=\int_0^c f'(x)\,dx-\int_c^1 f'(x)\,dx\\ &=f(c)-f(0)+f(c)-f(1)\\ &=2f(c). \end{aligned} \] Since \(f\) takes values in \([0,1]\), one has \(f(c)\le1\).
4. The arclength \(L\) of the graph of \(f\) satisfies\[ \boxed{L\le3}. \]
For every real number \(t\), \[ \sqrt{1+t^2}\le1+\abs t, \] because both sides are nonnegative and \[ 1+t^2\le(1+\abs t)^2. \] Therefore, using step 3, \[ \begin{aligned} L &=\int_0^1\sqrt{1+f'(x)^2}\,dx\\ &\le\int_0^1\left(1+\abs{f'(x)}\right)\,dx\\ &=1+\int_0^1\abs{f'(x)}\,dx\\ &\le3. \end{aligned} \]
5. Q.E.D.
Step 4 gives the required arclength bound.