KindiakMath

Bounded Derivative Integrals

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}

Hence,

\displaystyle \lim_{y \to \infty} \int_0^1 x^{y+1}\cdot f'(x)\, \mathrm dx = 0.

Since y/(y+1) \to 1 as y \to \infty, using limit laws,

\begin{aligned} \lim_{y \to \infty} \int_0^1 yx^y f(x)\, \mathrm dx &= 1 \cdot f(1) - 1 \cdot 0 = f(1). \end{aligned}

—Joel Kindiak, 11 Aug 25, 1118H

Problem 3. Suppose \displaystyle \lim_{T \to \infty} Tf(T) = 0. Evaluate \displaystyle \int_0^{\infty} tf'(t)\, \mathrm dt.

Solution. Integrating by parts,

\displaystyle \int_0^T tf'(t)\, \mathrm dt = [tf(t)]_0^T - \int_0^T f(t)\, \mathrm dt = Tf(T) - \int_0^T f(t)\, \mathrm dt.

We claim that \displaystyle \int_0^{\infty} f(t)\, \mathrm dt converges, so that

\displaystyle \int_0^{\infty} tf'(t)\, \mathrm dt = \lim_{T \to \infty} Tf(T) - \lim_{T \to \infty} \int_0^T f(t)\, \mathrm dt = \int_0^{\infty} f(t)\, \mathrm dt.

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joelkindiak

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