The Definition of Continuity

Prove that sums and products of continuous functions are continuous, and deduce that every polynomial is continuous.

\lvert f(x)g(x) - f(a)g(a)\rvert \leq \lvert f(x)\rvert\cdot\lvert g(x) - g(a)\rvert + \lvert g(a)\rvert\cdot\lvert f(x) - f(a)\rvert

a = 1

, the intact parabola — polynomials are continuous everywhere.

Show proof

Constants $c$ : $\lvert c - c\rvert = 0 < \varepsilon$ , so continuous. The identity $x \mapsto x$ : $\delta = \varepsilon$ works, since $\lvert x - a\rvert < \delta$ gives $\lvert x - a\rvert < \varepsilon$ .
(Sum) Let $f, g$ be continuous at $a$ . Given $\varepsilon > 0$ , take $\delta_1$ with $\lvert f(x) - f(a)\rvert < \frac{\varepsilon}{2}$ and $\delta_2$ with $\lvert g(x) - g(a)\rvert < \frac{\varepsilon}{2}$ ; with $\delta = \min(\delta_1, \delta_2)$ , $\lvert (f+g)(x) - (f+g)(a)\rvert \leq \lvert f(x) - f(a)\rvert + \lvert g(x) - g(a)\rvert < \varepsilon$ .
(Product) Use the identity $\lvert f(x)g(x) - f(a)g(a)\rvert \leq \lvert f(x)\rvert\cdot\lvert g(x) - g(a)\rvert + \lvert g(a)\rvert\cdot\lvert f(x) - f(a)\rvert$ . First bound $f$ near $a$ : by continuity of $f$ there is $\delta_0$ with $\lvert f(x) - f(a)\rvert < 1$ for $\lvert x - a\rvert < \delta_0$ , hence $\lvert f(x)\rvert \leq \lvert f(a)\rvert + 1 =: M$ there (local boundedness). Now take $\delta_1$ with $\lvert g(x) - g(a)\rvert < \frac{\varepsilon}{2M}$ and $\delta_2$ with $\lvert f(x) - f(a)\rvert < \frac{\varepsilon}{2(\lvert g(a)\rvert + 1)}$ ; with $\delta = \min(\delta_0, \delta_1, \delta_2)$ we get $\lvert f(x)g(x) - f(a)g(a)\rvert < M \cdot \frac{\varepsilon}{2M} + \lvert g(a)\rvert \cdot \frac{\varepsilon}{2(\lvert g(a)\rvert+1)} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ .
By induction $x^n$ is continuous (repeated product rule), $c x^n$ is continuous (constant times continuous), and their sum — a polynomial — is continuous (sum rule).

Prove $\lvert x\rvert$ is continuous at $0$ with $\delta = \varepsilon$ .

\left\lvert \lvert x\rvert - 0 \right\rvert = \lvert x\rvert < \varepsilon

a = 0

— continuity survives the cusp (differentiability is another story).

Show proof

We need $\lvert x - 0\rvert < \delta \Rightarrow \lvert \lvert x\rvert - \lvert 0\rvert \rvert < \varepsilon$ . But $\lvert \lvert x\rvert - 0\rvert = \lvert x\rvert$ , so it suffices to get $\lvert x\rvert < \varepsilon$ .
Put $\delta = \varepsilon$ : then $\lvert x - 0\rvert = \lvert x\rvert < \delta = \varepsilon$ is exactly the desired inequality. Hence $\lvert x\rvert$ is continuous at $0$ .
[framing] Continuity does not imply differentiability: $\lvert x\rvert$ is not differentiable at $0$ . Continuity is "no breaks", not "smoothness".

If $f$ is continuous at $a$ and $g$ is continuous at $f(a)$ , prove $h = g \circ f$ is continuous at $a$ .

\lvert x - a\rvert < \delta \Rightarrow \lvert f(x) - f(a)\rvert < \eta \Rightarrow \lvert g(f(x)) - g(f(a))\rvert < \varepsilon

a = -1

— continuity travels through composition: the chain

\varepsilon \leftarrow \eta \leftarrow \delta

.

Show proof

Let $\varepsilon > 0$ . Since $g$ is continuous at $b = f(a)$ , there is $\eta > 0$ with $\lvert g(y) - g(b)\rvert < \varepsilon$ whenever $\lvert y - b\rvert < \eta$ .
Now apply the continuity of $f$ to $\eta$ : there is $\delta > 0$ with $\lvert f(x) - f(a)\rvert < \eta$ whenever $\lvert x - a\rvert < \delta$ .
If $\lvert x - a\rvert < \delta$ then $\lvert f(x) - b\rvert < \eta$ , so $\lvert g(f(x)) - g(b)\rvert < \varepsilon$ , i.e. $\lvert h(x) - h(a)\rvert < \varepsilon$ . Hence $h$ is continuous at $a$ .
Numerically: if $g$ demands $\eta = \frac{\varepsilon}{2}$ and $f$ answers that $\eta$ with $\delta = \frac{\eta}{2}$ , the composition runs on $\delta = \frac{\varepsilon}{4}$ — the allowances compose along the chain.

The Definition of Continuity

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