Exercises – Page 4 – KindiakMath

In what follows, let $f : [0,1] \to \mathbb{R}$ be a continuously differentiable function. Suppose there exists $M > 0$ such that for any $x \in (0,1)$ , $|f'(x)| \leq M$ .

Problem 1. Prove that $\displaystyle f(1/2) \leq \frac{M}{4} + \int_0^1 f(x)\,\mathrm dx$ .

(Click for Solution)

Solution. The key trick is to notice that $[0, 1] = [0,1/2] \cup [1/2,1]$ . Analysing both parts then combining their results should yield our desired outcome.

Fix $x \in (1/2,1]$ . Since $f$ is differentiable, by the mean value theorem, there exists $c \in (1/2, x)$ such that

$f(x) = f(1/2) + f'(c)(x-1/2).$

Integrating on both sides yields

$\displaystyle \int_{1/2}^1 f(x)\, \mathrm dx = \int_{1/2}^1 f(1/2)\, \mathrm dx + \int_{1/2}^1 f'(c)(x-1/2)\, \mathrm dx.$

Since $f(1/2)$ is constant,

$\displaystyle \int_{1/2}^1 f(1/2)\, \mathrm dx = f(1/2) \int_{1/2}^1 1\, \mathrm dx = f(1/2) \cdot \frac 12.$

At this point in time it is tempting to bring out $f'(c)$ . However, $c$ depends on $x$ and therefore needs to remain in the integral. We will instead shift $f(1/2)$ to the left hand side apply $|\cdot|$ on all sides:

$\displaystyle \left| \int_{1/2}^1 f(x)\, \mathrm dx - f(1/2)\right| = \left|\int_{1/2}^1 f'(c)(x-1/2)\, \mathrm dx\right|.$

Applying the triangle inequality for integrals yields

$\displaystyle \begin{aligned} \left|\int_{1/2}^1 f'(c)(x-1/2)\, \mathrm dx\right| &\leq \int_{1/2}^1 |f'(c)| \cdot |x-1/2|\, \mathrm dx \\ &\leq \int_{1/2}^1 M \cdot (x-1/2)\, \mathrm dx \\ &= M \int_{1/2}^1 (x-1/2)\, \mathrm dx, \end{aligned}$

where we used $|f'(x)| \leq M$ in the second inequality, and $x > 1/2$ implies that $|x-1/2| = x-1/2 > 0$ . Evaluating the integral (either through antiderivatives or the area under a triangle) yields

$\displaystyle \int_{1/2}^1 (x-1/2)\, \mathrm dx = \frac 18.$

Consequently, combining the bounds, we obtain

$\displaystyle \left| \int_{1/2}^1 f(x)\,\mathrm dx - \frac 12 f(1/2) \right| \leq \frac{M}{8}.$

A similar argument on $[0,1/2]$ will yield

$\displaystyle \left| \int_0^{1/2} f(x)\,\mathrm dx - \frac 12 f(1/2) \right| \leq \frac{M}{8}.$

By the triangle inequality,

$\displaystyle \begin{aligned} \left| \int_0^{1} f(x)\,\mathrm dx - f(1/2) \right| &= \left| \int_{1/2}^1 f(x)\,\mathrm dx - \frac 12 f(1/2)+ \int_0^{1/2} f(x)\,\mathrm dx - \frac 12 f(1/2) \right| \\ &\leq \left| \int_{1/2}^1 f(x)\,\mathrm dx - \frac 12 f(1/2) \right| + \left| \int_0^{1/2} f(x)\,\mathrm dx - \frac 12 f(1/2) \right| \\ &\leq \frac M8 + \frac M8 = \frac M4,\end{aligned}$

as required.

Problem 2. Evaluate $\displaystyle \lim_{y \to \infty} \int_0^1 yx^y f(x)\, \mathrm dx$ in terms of $f(1)$ .

(Click for Solution)

Solution. Integrating by parts,

$\begin{aligned} \int_0^1 yx^y f(x)\, \mathrm dx &= \left[ y \cdot \frac {x^{y+1}}{y+1} \cdot f(x) \right]_0^1 - \int_0^1 y \cdot \frac {x^{y+1}}{y+1} \cdot f'(x) \, \mathrm dx \\ &= \frac y{y+1} \cdot f(1) - \frac y{y+1} \cdot \int_0^1 x^{y+1}\cdot f'(x)\, \mathrm dx. \end{aligned}$

Since $|f'(x)| \leq M$ , by the triangle inequality,

$\begin{aligned} \lim_{y \to \infty} \left| \int_0^1 x^{y+1}\cdot f'(x)\, \mathrm dx \right| & \leq \lim_{y \to \infty} \int_0^1 x^{y+1}\cdot |f'(x)|\, \mathrm dx \\ & \leq \int_0^1 x^{y+1}\cdot M\, \mathrm dx \\ &= M \cdot \lim_{y \to \infty} \frac{1}{y+2} = 0. \end{aligned}$