Let $f(x)$ be continuously differentiable twice on $[0,1]$ and $f(0)=f(1)=0$, and $|f''(x)| ≤A$ on $(0,1)$, prove $|f'(x)| ≤\frac{A}{2}$ on $[0,1]$.

I know thanks to Rolle's theorem that there exists a $c$ such that $f'(c)=0$, I'll use the mean value theorem on the derivative: $|\frac{f'(c)-f'(0)}{c}| ≤A$ and $|\frac{f'(1)-f'(c)}{1-c}| ≤A$

I'm not sure how this helps me.


1 Answer 1


You have to use the Lagrange mean value theorem more often than you have currently done. The point is that because $f'' \leq A$ uniformly, applying the Lagrange mean value theorem can give you a lot more information on the variation than merely applying it at the two extreme points.

For example, suppose that $a \neq b$ are arbitrary points in $[0,1]$. By the mean value theorem applied to $f'$ in the interval $[a,b]$ (or $[b,a]$), we get that $f'(b)-f'(a)= |b-a|f'(c)$ (or $f'(a)-f'(b) = |b-a|f'(c)$) for some $c \in (a,b)$ (or $(b,a)$), but because $|f''(c)| \leq A$ we have respectively $$-|b-a|A \leq f'(b) - f'(a) \leq |b-a|A$$ (or $$ -|b-a|A \leq f'(a) - f'(b) \leq |b-a|A $$ )Combining these, we get for all $a \neq b$ that $$ f'(a) - f'(b) \leq |b-a|A \\ f'(a) - f'(b) \geq -|b-a|A $$

In particular, let us now FIX an $a \in (0,1)$. Integrating the variable $b$ in the first inequality from $0$ to $1$ gives $$ \int_0^1 (f'(a)-f'(b))db \leq \int_{0}^1 |b-a|A db $$ the LHS is, by FTC $$ f'(a) - f(1) + f(0) = f'(a) $$ while the RHS is $$ A\int_0^1 |b-a|db = A \int_0^a (a-b)db + \int_a^1 (b-a)db = A\left(a^2-a+\frac 12\right) \leq \frac{A}{2} $$ because $a^2 -a\leq 0$ if $a \in [0,1]$. Thus, we are led to $f'(a) \leq \frac{A}{2}$.

By similarly integrating the bound $f'(a) - f'(b) \geq -|b-a|A$ with respect to $b$ from $0$ to $1$, we get $f'(a) \geq -\frac{A}{2}$, leading to the answer, because $a$ was arbitrary in $[0,1]$, although fixed for the argument above.


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