# Prove that the product of two continuous functions is continuous

I was trying to prove an important theorem concerning continuous functions, namely that the product of two continuous functions is continuous. I am nearly at the end of the proof but I do not understand the last step my teacher made... I'm sure some of you is able to help me out :) Thanks in advance!

Theorem

Let $f, g\colon \Bbb R \to\Bbb R$ be continuous. Then $F\colon \Bbb R\to\Bbb R$ defined by $F(x) = f(x)g(x)$ is continuous.

Proof

Let $f, g\colon \Bbb R \to\Bbb R$ be given such that $f$ and $g$ are continuous.

Let $F\colon \Bbb R\to\Bbb R$ be defined by $F(x) = f(x)g(x)$. We want to show: $F$ is continuous, that is, for all $a \in\Bbb R$, for every $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in \Bbb R$ with $|x - a| < \delta$, $|F(x)-F(a)| < \epsilon$.

Let $a \in\Bbb R$ be given. Let $\epsilon > 0$ be given.

Let $\delta_f > 0$ be such that for all $x \in\Bbb R$ with $|x-a| < \delta_f$, $|f(x)-f(a)| < \frac\epsilon{2(|g(a)| + \epsilon)}$. Such a $\delta_f$ exists since $f$ is continuous.

Let $\delta_g > 0$ be such that for all $x \in\Bbb R$ with $|x-a| < \delta_g$, $|g(x)-g(a)| < \frac\epsilon{ 2|f(a)| + 1}$. Such a $\delta_g$ exists since $g$ is continuous.

Let $\delta_3 > 0$ be such that for all $x\in\Bbb R$ with $|x-a| < \delta_3$, $|g(x)-g(a)| < \epsilon$, which implies $|g(x)| < |g(a)| + \epsilon$. Such a $\delta_3$ exists since $g$ is continuous.

Choose $\delta = \min\{\delta_f, \delta_g, \delta_3\}$.

Then for all $x \in\Bbb R$ with $|x-a| < \delta$: \begin{align}|F(x) - F(a)| &= |f(x)g(x) - f(a)g(a)| \\ &= |f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)| \\ &\le |f(x)g(x) - f(a)g(x)| + |f(a)g(x) - f(a)g(a)|\\ &= |(g(x)[f(x) - f(a)]| + |f(a)[g(x) - g(a)]|\\&= |g(x)| * |f(x)-f(a)| + |f(a)| * |g(x) - g(a)| \\ &< [|g(x)| + \epsilon] \cdot \frac \epsilon {2(|g(a)| + \epsilon)} + |f(a)| \cdot |g(x) - g(a)| \\ &= \frac\epsilon{2 + |f(a)| \cdot |g(x) - g(a)|}\end{align}

Problem

Now we need $|f(a)| \cdot |g(x) - g(a)|$ to be equal to $\frac \epsilon 2$ so that we have $\frac\epsilon2+\frac\epsilon 2= \epsilon$ (which completes the proof). The problem is I do not see why/how $|f(a)| \cdot |g(x) - g(a)|$ is equal to $\frac\epsilon2$. Or is there maybe an error somewhere?

• – ً ً Dec 30 '15 at 11:58
• you don't need $|f(a)||g(x)-g(a)|=\frac{\epsilon}{2}$, you only need it to be less than $\frac{\epsilon}{2}$, and your choice above makes that true. – Arun Kumar Dec 30 '15 at 12:16

You've set up $\delta_f,\delta_g$ and $\delta_3$ correctly. $\forall x \in \Bbb R$ with $|x-a|<\delta$,

\begin{align}|F(x) - F(a)| &= |f(x)g(x) - f(a)g(a)| \\ &= |f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)| \\ &\le |f(x)g(x) - f(a)g(x)| + |f(a)g(x) - f(a)g(a)|\\ &= |(g(x)[f(x) - f(a)]| + |f(a)[g(x) - g(a)]|\\&= |g(x)| * |f(x)-f(a)| + |f(a)| * |g(x) - g(a)| \\ &< \color{red}{[|g(a)| + \epsilon]} \cdot \frac \epsilon {2(|g(a)| + \epsilon)} + |f(a)| \cdot |g(x) - g(a)| \\ &\le \frac\epsilon{2}+|f(a)|\cdot \frac\epsilon{ 2|f(a)| + 1}\\ &< \frac\epsilon{2}+(|f(a)|+\frac12)\cdot \frac\epsilon{ 2|f(a)| + 1}\\ &= \frac\epsilon{2}+\frac\epsilon{2}=\epsilon\end{align}

The red part holds because $\forall x \in \Bbb R$ with $|x-a|<\delta<\delta_3$, $|g(x)-g(a)| < \epsilon \implies |g(x)| \le |g(x)-g(a)| + |g(a)| < |g(a)| + \epsilon$.

from given , i can conclude $(|f(a)| + \frac{1}{2})|g(x) - g(a)| < \frac{\epsilon}{2}$ $\Rightarrow |f(a)|*|g(x) - g(a)| < (|f(a)| + \frac{1}{2})|g(x) - g(a)|$ $\Rightarrow |f(a)|*|g(x) - g(a)| < \frac{\epsilon}{2}$. I think this was your doubt.

$|g(x)|<|g(a)|+\epsilon=B'$

$B=max(B', |f(a)|)$

$|g(x) - g(a)|<\epsilon/(2B)$

$|f(x)-f(a)|<\epsilon/(2B)$

Whenever, $|x-a|<\delta$ $\implies$ $|F(x) - F(a)| =|g(x)||f(x) - f(a)|+|f(a)||g(x)-g(a) | < B'*\epsilon/(2B)+|f(a)|*\epsilon/(2B) <B*\epsilon/(2B)+B*\epsilon/(2B)<\epsilon/2+\epsilon/2=\epsilon$ (proved)

• Please use MathJax in your posts. – Kamil Jarosz Dec 30 '15 at 13:17