KindiakMath

Category: Calculus

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

October 18, 2024

Calculus, math, mathematics
Exploring o-Calculus

I was inspired by this post on Knuth’s approach to using $o$ -notation to simplify limits. It turns out that independently, I explored a similar variant in this writeup. For a first post in this blog, perhaps it is wise to revisit this notion, but now cleaning it up using $o$ -notation.

We will first define $o(1)$ to have what we conventionally mean by a function with limit $0$ as $x \to 0$ .

Definition 1. Let $f$ be a real-valued function defined on the interval $(-r,r) \backslash \{0\}$ for some $r > 0$ . We write $f \in o(1)$ to mean:

For any $\epsilon > 0$ , there exists $\delta > 0$ such that $0 < |t| < \delta$ implies $|f(t)| < \epsilon$ .

Intuitively, this definition means that for any output error threshold $\epsilon$ , there exists an input error threshold $\delta$ such that inputs $t \approx 0$ (i.e. $0 < |t| < \delta$ ) yield outputs $f(t) \approx 0$ (i.e. $|f(t)| < \epsilon$ ).

Indeed, conventionally this is equivalent to the notation $\displaystyle \lim_{x \to 0} f(x) = 0$ . Furthermore, we will denote $f(x) = o(1)$ to mean $f \in o(1)$ and

$f(x) = g(x) + o(1)\quad \iff \quad f(x) - g(x) = o(1).$

Using standard $\epsilon$ – $\delta$ arguments, we can prove the following limit laws:

Theorem 1. The following properties hold: for any real $k$ ,

$\begin{aligned} o(1) + o(1) &= o(1), \\ k \cdot o(1) &= o(1), \\ o(1) \cdot o(1) &= o(1).\end{aligned}$

Proof. Fix $f \in o(1)$ and $g \in o(1)$ and $\epsilon > 0$ . Then, for any $k_i > 0$ , there exists $\delta_i > 0$ such that $0 < |t| < \min\{\delta_1,\delta_2\} =: \delta$ implies

$|f(t)| < k_1 \cdot \epsilon \quad \text{and} \quad |g(t)| < k_2 \cdot \epsilon$ .

For the first result, the triangle inequality yields

$|(f+g)(t)| \leq |f(t)| + |g(t)| < (k_1 + k_2) \cdot \epsilon,$

so that setting $k_1 = k_2 = 1/2$ yields $f+g \in o(1)$ .

For the second result, set $k_1 = 1$ if $k = 0$ . Otherwise,

$|(kf)(t)| = |k||f(t)| < |k| \cdot k_1 \cdot \epsilon$

so that setting $k_1 = 1/|k|$ yields $kf \in o(1)$ .

For the third result,

$|(fg)(t)| = |f(t)||g(t)| < k_1 \cdot k_2 \cdot \epsilon^2$

so that setting $k_1 = 1$ and $k_2 = 1/\epsilon$ yields $fg \in o(1)$ .

With a little bit more flexibility, we can also prove an analogous result for division, but more on that later. Many elementary limit-based results can be derived similarly, such as the squeeze theorem.

Theorem 2 (Squeeze Theorem). Let $f,g,h$ be real-valued functions defined on $(-r,r)\backslash \{0\}$ for some $r > 0$ such that $f \leq g \leq h$ . If $f,h \in o(1)$ , then $g \in o(1)$ .

Proof Sketch. Perform epsilontics to derive

$-k_1 \cdot \epsilon < f(x) \leq g(x) \leq h(x) < k_2 \cdot \epsilon$

for suitably chosen $\delta_1,\delta_2 > 0$ . Setting $k_1 = k_2 = 1$ yields the desired result.

The power of analysing $o(1)$ functions arises in defining and proving the more general limits used in practice.

Definition 2. Let $f$ be a real-valued function defined on the interval $(-r,r) \backslash \{0\}$ for some $r > 0$ . We write $\displaystyle \lim_{x \to 0} f(x) = L$ to mean $f(x) = L + o(1)$ .

We write $f(x) = g(o(1))$ to mean that there exists $h = o(1)$ such that for any $x \in (-r,r) \backslash \{0\}$ for some $r > 0$ , $f(x) = g(h(x))$ .

Theorem 3. $\displaystyle \frac{1}{1+ o(1)} = 1 + o(1)$ .

Proof. Let $\displaystyle f(x) = \frac{1}{1+g(x)}$ for some $g \in o(1)$ . Fix $\epsilon > 0$ . It suffices to show that $\displaystyle f(x) - 1 = o(1)$ . To that end, consider the bound

$\displaystyle |f(x) - 1| = \left|\frac{1}{1+g(x)}-1\right| = \frac{1}{|1+g(x)|} \cdot |g(x)|.$

We take advantage of $g \in o(1)$ in two ways. For any $k_i$ , $i=1,2$ , there exists $\delta_i$ such that

$0 < |x| <\delta_i \quad \Rightarrow \quad |g(x)| < k_i \cdot \epsilon.$

For $k_1$ , we have the bound $-k_1 \cdot \epsilon < g(x) < k_1 \cdot \epsilon$ implying, if $1-k_1 \cdot \epsilon > 0$ , that

$\displaystyle \frac{1}{1+g(x)} < \frac{1}{1-k_1 \cdot \epsilon}.$

Therefore, defining $\delta := \min\{\delta_1,\delta_2\}$ , the complete bound is given by

$\displaystyle |f(x) - 1| = \frac{1}{|1+g(x)|} \cdot |g(x)| < \frac{1}{1-k_1 \cdot \epsilon} \cdot k_2 \cdot \epsilon.$

To complete the proof, we then simply choose $k_1 = 1/(2\epsilon)$ and $k_2 = 1/2$ . Thus, $f(x) - 1 \in o(1)$ , as required.

The limit laws for such functions then generalise rather naturally. We illustrate using the multiplicativity of taking limits.

Theorem 4. Let $f,g$ be real-valued functions defined on the interval $(-r,r) \backslash \{0\}$ for some $r > 0$ . Suppose $\displaystyle \lim_{x \to 0} f(x) = L$ and $\displaystyle \lim_{x \to 0} g(x) = M$ . Then $\displaystyle \lim_{x \to 0} (f(x)g(x)) = LM$ .

Proof. Write $f(x) = L+ o(1)$ and $g(x) = M+o(1)$ . Then

$\begin{aligned} f(x)g(x) &= (L+o(1))(M+o(1)) \\ &= LM + L\cdot o(1) + M \cdot o(1) + o(1) \cdot o(1) \\ &= LM + o(1) + o(1) + o(1) \\ &= LM + o(1).\end{aligned}$

Therefore, $\displaystyle \lim_{x \to 0} (f(x)g(x)) = LM$ .

We can even generalise to limits at any real $c$ .

Definition 3. Let $c$ be a real number and $f$ be a real-valued function defined on $(c-r,c+r) \backslash \{0\}$ for some $r > 0$ . We write $\displaystyle \lim_{x \to c} f(x) = L$ to mean $\displaystyle \lim_{t \to 0} f(c+t) = L$ . Equivalently, $f(c + o(1)) = L + o(1)$ .

One can check that this does indeed satisfy the usual $\epsilon$ – $\delta$ definition for limits. We can therefore define (local) continuity and differentiability. For our next post, we will use the local definition of differentiability to compute the derivative of $x^n$ .

—Joel Kindiak, 18 Oct 24, 0745H

October 18, 2024

math, mathematics