how to prove that given f(x) and g(x) is continuous then $f*g(x)$ is also continuous

Question

Question

how to prove that given f(x) and g(x) is continuous then $f*g(x)$ is also continuous

All I have is that because we already know that f(x) and g(x) is continuous then $|x-a|<\delta \implies |f(x)-f(a)| $ and $|g(x)-g(a)|<\epsilon$

And I also know that

$\lim{_{x\to a}}{f(x)}g(x)=\lim{_{x\to a}}{g(x)}*\lim{_{x\to a}}{f(x)}=f(a)*g(a)$ but I don't know what to do with those information

Answer 1

This can be the 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 $\delta = \min\{\delta_f, \delta_g, \delta_3\}$.

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

Answer 2

I think $*$ means product.

Hint:

$|f(x)g(x)-f(a)g(a)|=|f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)| \le |f(x)||g(x)-g(a)|+|g(a)|f(x)-f(a)|.$

Answer 3

There are a number of ways to do this. I always like to use sequences since they tend to simplify things. Let $x_n \to x$. Then, \begin{align} |(fg)(x_n) - (fg)(x)| &= |f(x_n)g(x_n) - f(x)g(x)|\\ &= | f(x_n)g(x_n) - f(x)g_n(x) + f(x)g_n(x) - f(x)g(x)|\\ &= | g(x_n)(f(x_n) - f(x)) + f(x)(g_n(x) - g(x))|\\ &\le |g(x_n)||f(x_n) - f(x)| + |f(x)||g_n(x) - g(x)|\\ &= |g(x_n) - g(x) + g(x)||f(x_n) - f(x)| + |f(x)||g_n(x) - g(x)|\\ &\le (|g(x_n) - g(x)| + |g(x)|)|f(x_n) - f(x)| + |f(x)||g_n(x) - g(x)| \end{align} Then by continuity of $f$ and $g$, we can make $|f(x_n) - f(x)| \to 0$ and $|g_n(x) - g(x)| \to 0$ so that the result follows. If you want to use an $\epsilon$-$\delta$ argument, the proof is very similar.