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Mathematics

real analysis

3 pages

01

Continuity and Equivalent Definitions

Definition (ε\varepsilon-δ\delta)

A function f:A→Rf:A\to\mathbb{R} is continuous at a∈Aa\in A if

∀ ε>0, ∃ δ>0such that∣x−a∣<δ  ⟹  ∣f(x)−f(a)∣<ε.\forall\,\varepsilon>0,\ \exists\,\delta>0 \quad\text{such that}\quad |x-a|<\delta \implies |f(x)-f(a)|<\varepsilon.

Idea: No matter how small a target tube ε\varepsilon around f(a)f(a) you demand, you can find an input window δ\delta around aa that lands inside it.

Equivalent Definitions

Limit form: lim⁡x→af(x)=f(a).\displaystyle\lim_{x\to a} f(x) = f(a).

Sequential form: for every sequence (xn)(x_n),

xn→a  ⟹  f(xn)→f(a).x_n \to a \implies f(x_n) \to f(a).

Note: The sequential form is often the easiest to disprove continuity — just exhibit one sequence xn→ax_n \to a with f(xn)↛f(a)f(x_n)\not\to f(a).

Example

Take f(x)=x2f(x) = x^2. For any aa,

lim⁡x→ax2=a2=f(a),\lim_{x\to a} x^2 = a^2 = f(a),

so f∈C(R)f \in C(\mathbb{R}). Likewise g(x)=1xg(x)=\dfrac{1}{x} satisfies lim⁡x→a1x=1a=g(a)\displaystyle\lim_{x\to a}\frac{1}{x}=\frac{1}{a}=g(a) for a>0a>0, hence g∈C((0,∞))g\in C\big((0,\infty)\big). ✓\checkmark

Problem

Let f:[0,1]→Rf:[0,1]\to\mathbb{R} satisfy: (1) ff has the Intermediate Value Property, and (2) for every c∈Rc\in\mathbb{R}, the level set f−1(c)f^{-1}(c) is closed. Prove ff is continuous on [0,1][0,1].

Solution

Setup. Suppose, for contradiction, ff is not continuous at some x0∈[0,1]x_0\in[0,1]. Then there is a sequence with

xn→x0,f(xn)↛f(x0).x_n \to x_0, \qquad f(x_n)\not\to f(x_0).

Passing to a subsequence, there exists ε>0\varepsilon>0 with

f(xn)≥f(x0)+ε.f(x_n) \geq f(x_0)+\varepsilon.

Apply IVP. Set α=f(x0)+ε2\alpha = f(x_0)+\dfrac{\varepsilon}{2}, so that f(x0)<α<f(xn)f(x_0) < \alpha < f(x_n). By the Intermediate Value Property, there is yny_n between x0x_0 and xnx_n with

f(yn)=α.f(y_n) = \alpha.

Since ∣yn−x0∣≤∣xn−x0∣|y_n - x_0| \leq |x_n - x_0|, we get yn→x0y_n \to x_0.

Use closedness. Let A=f−1(α)A = f^{-1}(\alpha). Each yn∈Ay_n \in A and yn→x0y_n \to x_0; as AA is closed,

x0∈A  ⟹  f(x0)=α=f(x0)+ε2,x_0 \in A \implies f(x_0) = \alpha = f(x_0)+\frac{\varepsilon}{2},

a contradiction. Therefore ff is continuous on [0,1][0,1].

f∈C([0,1])\boxed{f \in C([0,1])}

02

Heine–Cantor Theorem

Heine--Cantor Theorem

Let II be a compact interval. Let f:I→Rf : I \to \mathbb{R} be continuous on II. Then ff is uniformly continuous on II.

Example

Let f(x)=x2,f(x)=x^2, on I=[−1,1].I=[-1,1]. The interval [−1,1][-1,1] is compact. The function ff is continuous on [−1,1][-1,1]. Therefore, f(x)=x2f(x)=x^2 is uniformly continuous on [−1,1][-1,1].

03

Weierstrass Inequality

Weierstrass Inequality

Statement

Let a1,a2,…,an∈[0,1)a_1, a_2, \dots, a_n \in [0,1). Then:

∏k=1n(1−ak)≥1−∑k=1nak\prod_{k=1}^{n} (1 - a_k) \geq 1 - \sum_{k=1}^{n} a_k

Similarly, for a1,a2,…,an≥0a_1, a_2, \dots, a_n \geq 0:

∏k=1n(1+ak)≥1+∑k=1nak\prod_{k=1}^{n} (1 + a_k) \geq 1 + \sum_{k=1}^{n} a_k

Idea: Multiplying the factors (1±ak)(1 \pm a_k) produces all the cross terms aiaja_i a_j. Keeping only the linear part gives the bound; the discarded terms control the sign.

Proof (by induction)

Base case n=1n = 1: equality holds, since 1−a1=1−a11 - a_1 = 1 - a_1.

Inductive step: Assume ∏k=1n(1−ak)≥1−∑k=1nak\displaystyle\prod_{k=1}^{n}(1 - a_k) \geq 1 - \sum_{k=1}^{n} a_k. Then:

∏k=1n+1(1−ak)≥(1−∑k=1nak)(1−an+1)\prod_{k=1}^{n+1}(1 - a_k) \geq \left(1 - \sum_{k=1}^{n} a_k\right)(1 - a_{n+1})

Expanding the right-hand side:

=1−∑k=1n+1ak+an+1∑k=1nak≥1−∑k=1n+1ak= 1 - \sum_{k=1}^{n+1} a_k + a_{n+1}\sum_{k=1}^{n} a_k \geq 1 - \sum_{k=1}^{n+1} a_k

since the extra term an+1∑k=1nak≥0a_{n+1}\sum_{k=1}^{n} a_k \geq 0. ■\blacksquare

Example

Take ak=12ka_k = \dfrac{1}{2^k} for k=1,2,3k = 1, 2, 3. Then ∑k=13ak=12+14+18=78\displaystyle\sum_{k=1}^{3} a_k = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}, so the inequality predicts:

∏k=13(1−12k)≥1−78=18\prod_{k=1}^{3}\left(1 - \frac{1}{2^k}\right) \geq 1 - \frac{7}{8} = \frac{1}{8}

Checking the exact product:

12⋅34⋅78=2164≈0.328  ≥  18=0.125✓\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} = \frac{21}{64} \approx 0.328 \;\geq\; \frac{1}{8} = 0.125 \quad \checkmark

Elegant takeaway: Even an infinite product stays positive — since ∑12k=1<∞\sum \frac{1}{2^k} = 1 < \infty, the bound guarantees ∏k=1∞(1−12k)>0\prod_{k=1}^{\infty}\left(1 - \frac{1}{2^k}\right) > 0.