Prove an inequality for the absolute value of derivatives of a function

Let $x \in [0,a]$ and assume $f$ is a function with $f''$ continuous in $[0,a]$ such that

$| f''(x)| \leq m.$

Furthermore, assume $f(x)$ is maximal at a point $x \in (0,a)$ (i.e., $f(x)$ does not have its maximum at either endpoint of the interval). Prove that

$|f'(0)| + |f'(a)| \leq am.$

Proof. Since $f(x)$ attains its maximum on the interval $(0,a)$ we know there is some $c \in (0,a)$ such that $f'(c) = 0$ . Then,

$\int_0^a f''(x) \, dx = \int_0^c f''(x) \, dx + \int_c^a f''(x) \, dx.$

Evaluating these integrals separately, we have (by the first fundamental theorem of calculus, which is permissible since $f''$ is continuous by hypothesis)

$\int_0^c f''(x) \, dx = f'(c) - f'(0) = - f'(0) \quad \implies \quad |f'(0)| = \left| \int_0^c f''(x) \, dx \right|.$

Now, we use the bound $f''(x) \leq m$ for all $x \in [0,a]$ ,

$|f'(0)| = \left| \int_0^c f''(x) \, dx \right| \leq \int_0^c |f''(x)| \, dx \leq \int_0^c m \, dx = mc.$

Next, we evaluate the second integral,

$\int_c^a f''(x) \, dx = f'(a) - f'(c) = f'(a).$

And so,

$|f'(a)| =\left| \int_c^a f''(x) \, dx \right| \leq \int_c^a |f''(x)| \, dx \leq m(a-c).$

Therefore,

$|f'(0)| + |f'(a)| \leq mc + m(a-c) = ma. \qquad \blacksquare$

Stumbling Robot

Prove an inequality for the absolute value of derivatives of a function

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