Home » Bounds

Tag: Bounds

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,

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


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