# Proof verification: the product of two continuous functions is continuous

Prove that if $f, g$ : $X$ → $\mathbb R$ are continuous at $a$ ∈ $\mathbb R$, then $f · g$ is continuous in $a$.

If $f$ is continuous at $a$, then $∀ε_f > 0, ∃δ_f > 0$ such that

$|x-a| < δ_f$ iff $|f(x) - f(a)| < ε_f$

and if $g$ is continuous at $a$, then $∀ε_g > 0, ∃δ_g > 0$ such that

$|x-a| < δ_g$ iff $|g(x) - g(a)| < ε_g$.

Now make $|f(x)g(x) - f(a)g(a)|$ = $|f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)|$ $≤$ $|f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)|$ < $ε_f|g(a)| + ε_g|f(x)|$. (1)

Notice that if $|f(x) - f(a)| < ε_f$ then $|f(x)| - |f(a)| < ε_f$, so $|f(x)| < ε_f + |f(a)|$.

That implies, from (1): $|f(x)g(x) - f(a)g(a)| < ε_f|g(a)| + ε_g(ε_f + |f(a)|) = ε_f(|g(a)| + ε_g|f(a)|)$.

Now I'm a little lost. From the proofs I've seen, it seems to me that I could simply take $δ = min(δ_g, δ_f)$. Also, since $ε_f$ and $ε_g$ can be made as small as one wants, there will be some δ that satisfies the conditions for continuity. But from what I read, I should explicitly show a δ in terms of ε. Anyway, I reasoned that since $f$ and $g$ are continuous, I can make bounds on $x$ symmetrically* as done to $f(x)g(x)$ and got the following:

δ = $δ_f(|a| + δ_g|a|)$ = $δ_f|a|(1 + δ_g)$

so:

$|x-a| < δ_f|a|(1 + δ_g)$ iff $|f(x)g(x) - f(a)g(a)| < ε_f(|g(a)| + ε_g|f(a)|)$

• I need to clarify/better define this, but it is basically the notion that making the same operations I made on the bounds of $f(x)g(x)$ in relation to the bounds of $f(x)$ and $g(x)$ individually over the bounds of $x$ for each function, should leave me with the adequate bounds for $x$ on the composite function.
• One could first have a look at the case of limit of sequences: if $\lim_{n\to\infty}s_n=s$, $\lim_{n\to\infty}t_n=t$, then $\lim_{n\to\infty}s_{n}t_{n}=st$. This follows from the equality $s_{n}t_{n}-st=(s_n-s)(t_n-t)+s(t_n-t)+t(s_n-s)$. Commented Aug 24, 2016 at 4:57
• I think using the sequences definition of continuity is the neatest way of solving this question as well. Commented Aug 24, 2016 at 5:11
• I need to point out you claim $|x-y| < \delta$ iff $|f(x)-f (y)|<\epsilon$. This is certainly **not** true. $|x-y|<\delta \implies |f (x)-f (y)|< \epsilon$ but it does not go they other way. Commented Aug 24, 2016 at 7:40

You seem to have the foundations of a correct proof. Here are some things to think about when doing proofs of real analysis. Essentially you want to prove that, given any $$ε > 0$$, there exists some $$\delta>0$$ s.t. $$|x-a| < \delta \rightarrow |f(x)g(x) - f(a)g(a)| < ε$$.

So let us pick an $$ε$$. We need to prove there is some $$\delta$$ that can satisfy the aforementioned property. Now before we find this $$\delta$$ we need to investigate how we can get it. Looking at your (1) below:

$$\begin{split} &|f(x)g(x) - f(a)g(a)| \\ &=|f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)| \\ &≤ |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)| \\ &< ε_f|g(a)| + ε_g|f(x)|. \end{split}$$ Now if you can pick a $$\delta$$ such that $$ε_f|g(a)| + ε_g|f(x)| \leq ε$$ then we are done. Our aim will be to get $$ε_f|g(a)| \leq \frac{ε}{2}$$ and $$ε_g|f(x)| \leq \frac{ε}{2}$$ so their sum will be less than or equal to $$ε$$.

It is important to remember that, because of the continuity of f and g, you have the choice to pick $$ε_f$$ and $$ε_g$$ to have whatever values you want.

So let $$\delta_f$$ be the appropriate value such that $$ε_f = \frac{ε}{2|g(a)|+1}$$ (2) and thus $$ε_f|g(a)| = \frac{ε|g(a)|}{2|g(a)|+1}< \frac{ε}{2}$$ (the +1 in the denominator is there to avoid division by 0).

Picking $$ε_g$$ is harder because we need $$ε_g|f(x)| \leq \frac{ε}{2}$$ and the $$f(x)$$ term is not a constant like $$g(a)$$ was in the previous case. We need to bound the $$f(x)$$ somehow. Well since f is continuous, if we let $$x$$ and $$a$$ be close enough to each other, we can bound f. Let us pick a $$\delta_b$$ s.t. for $$|x-a|<\delta_b$$, we have $$|f(x) - f(a)| < ε \implies |f(x)| < ε + |f(a)|$$ (3) by triangle ineq. And so we have, for $$x$$ and $$a$$ close enough, $$ε_g|f(x)| < ε_g(ε + |f(a)|)$$ and so we let $$\delta_g$$ be the appropriate value such that $$ε_g = \frac{ε}{2(ε +|f(a)|)}$$ (4) and so $$ε_g|f(x)| < \frac{ε}{2(ε +|f(a)|)}(ε + |f(a)|) = \frac{ε}{2}$$

Thus given that $$x$$ and $$a$$ are close enough to each other (explained at the end), we can get $$ε_f|g(a)| \leq \frac{ε}{2}$$ and $$ε_g|f(x)| \leq \frac{ε}{2}$$ and so $$ε_f|g(a)| + ε_g|f(x)| \leq ε$$ and so

$$\begin{split} &|f(x)g(x) - f(a)g(a)| < ε_f|g(a)| + ε_g|f(x)| \leq ε \\ &\implies |f(x)g(x) - f(a)g(a)| < ε \end{split}$$ as required.

But what does it mean for $$x$$ and $$a$$ to be close enough? We need to specify how close they actually need to be (this is ultimately our $$\delta$$ that we are trying to find). Well we need $$|x-a| < \delta_f$$ for (2) and we need $$|x-a| < \delta_b$$ for (3) and we need $$|x-a| < \delta_g$$ for (4) and so we can say $$x$$ and $$a$$ are close enough if $$|x-a| < min\{\delta_f,\delta_g,\delta_b\}$$ and so $$\delta=min\{\delta_f,\delta_g,\delta_b\}$$

I hope this helps.

• This is a brilliant proof. Anyone new to epsilon-delta proofs can test their conceptual understanding with this answer Commented Oct 15, 2019 at 10:05
• This is such a great answer. Note that since $|a| = |(a-b) + b| \leq |a-b| + |b|$ then $|a| - |b| \leq |a-b|$. Thus we have that $|f(x)| - |f(a)| \leq |f(x) - f(a)| < \epsilon$ giving us (as stated above) $|f(x)| < \epsilon + |f(a)|$. Commented Dec 30, 2020 at 21:43
• nice It should be saved Commented Jun 18, 2021 at 22:47
• This is insane!!! Note that $X$ could be an arbitrary topological space. So the "required $\delta$" is the intersection of the three balls "defined by the $\delta_f, \delta_g, \delta_b$". Commented Apr 19 at 2:17

The essence of a continuity proof is to show that for any $\epsilon$, you can find a $\delta$, and usually those proofs are constructive (you indeed establish a formula for $\delta$ as a function of $\epsilon$).

In the case at hand, you know that such a relation holds for $f$ and $g$ and need to establish it for $f\cdot g$. Specifically,

$$\forall\epsilon_f,\epsilon_g:\exists \delta_f,\delta_g\implies\forall\epsilon_{f\cdot g}:\exists\delta_{f\cdot g}.$$

The next step is to show that for an arbitrary $\epsilon_{f\cdot g}$ you can choose values of $\epsilon_f,\epsilon_g$ such that the continuity condition holds for $f\cdot g$ (see Appendix). Then by continuity of $f$ and $g$, the global continuity condition holds when you are inside both corresponding $\delta_f,\delta_g$ neighborhoods of $a$, i.e. in a neighborhood of radius $\min(\delta_f,\delta_g)$.

$$\epsilon_{f\cdot g}\xrightarrow[\text{assignment}]{}\epsilon_f,\epsilon_g\xrightarrow[\text{confinuity of }f,g]{}\delta_f,\delta_g\xrightarrow[\text{common neighborhood}]{}\delta_{f\cdot g}.$$

This establishes a functional relation between $\epsilon_{f\cdot g}$ and $\delta_{f\cdot g}$.

Appendix:

As you established,

$$|f(x)g(x)-f(a)g(a)|<|f(x)|\,|g(x)-g(a)|+|f(x)-f(a)|\,|g(a)|$$which annoyingly involves $f(x)$ and is better replaced by

$$<|f(a)|\,|g(x)-g(a)|+|f(x)-f(a)|\,|g(a)|+|f(x)-f(a)|\,|g(x)-g(a)|.$$

In terms of the $\epsilon$,

$$|f(x)g(x)-f(a)g(a)|<|f(a)|\epsilon_g+|g(a)|\epsilon_f+\epsilon_f\epsilon_g<\epsilon.$$

To achieve the last inequality, you are free to define the $\epsilon$ in a way that suits you, for example by ensuring that none of the terms exceeds a third of $\epsilon$: $$\epsilon_f<\min\left(\frac{\epsilon_{f\cdot g}}{3|g(a)|},\sqrt{\frac{\epsilon_{f\cdot g}}3}\right),\\ \epsilon_g<\min\left(\frac{\epsilon_{f\cdot g}}{3|f(a)|},\sqrt{\frac{\epsilon_{f\cdot g}}3}\right).$$

After you have established that $$|f(x)g(x)-f(a)g(a)|\le|f(x)g(x)-f(a)g(x)|+|f(a)g(x)-f(a)g(a)|\\ =|f(x)-f(a)||g(x)|+|f(a)||g(x)-g(a)|,$$

by continuity of $$g(x)$$, you can make the factor $$|g(x)|$$ arbitrarily close to $$|g(a)|$$, and it is not a big deal to find $$\delta_f,\delta_g$$ that ensure

$$\epsilon_f(|g(a)|+\epsilon_g)+|f(a)|\epsilon_g<\epsilon.$$

Let $$f(x)-f(a)=u(x,a)=u$$ and $$g(x)-g(a)=v(x,a)=v$$ and $$f(x)g(x)-f(a)g(a)=w(x,a)=w$$ $$w(x,a)=f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)=f(x)v(x,a)+g(a)u(x,a)=(u(x,a)+f(a))v(x,a)+g(a)u(x,a)=uv+f(a)v+g(a)u$$ and hence $$|w| \leq |uv|+|f(a)||v|+|g(a)||u|$$ assume first $$f(a)$$ and $$g(a)$$ are both nonzero. in that case divide both sides by $$|f(a)g(a)|$$

$$\frac{|w|}{|f(a)*g(a)|} \leq \frac{|u|}{|f(a)|}*\frac{|v|}{|g(a)|}+\frac{|v|}{|g(a)|}+\frac{|u|}{|f(a)|} =(1+\frac{|u|}{|f(a)|})(1+\frac{|v|}{|g(a)|})-1$$

so that $$1+\frac{|w|}{|f(a)*g(a)|} \leq (1+\frac{|u|}{|f(a)|})*(1+\frac{|v|}{|g(a)|})$$ we would like $$|w|< \epsilon$$ let $$\epsilon_f \leq |f(a)|*\left(\sqrt[2]{1+\frac{\epsilon}{|f(a)*g(a)|}}-1\right)$$ and $$\epsilon_g \leq |g(a)|*\left(\sqrt[2]{1+\frac{\epsilon}{|f(a)*g(a)|}}-1\right)$$ in other words choose $$\delta = \min(\delta_f,\delta_g)$$ where $$\delta_f$$ and $$\delta_g$$ correspond to $$\epsilon_f$$ and $$\epsilon_g$$ with above choice. Then $$|u| < \epsilon_f$$ and $$|v| < \epsilon_g$$ will imply $$|w(x,a)| < \epsilon$$

In case $$f(a)=0$$ and $$g(a) \neq 0$$ then choose $$\epsilon_f \leq \min( \sqrt{\frac{\epsilon}{2}},\frac{\epsilon}{2|g(a)|})$$ and $$\epsilon_g \leq \sqrt{\frac{\epsilon}{2}}$$ and then for corresponding $$\delta_f$$ and $$\delta_g$$ let $$\delta=\min(\delta_f,\delta_g)$$. If $$|x-a|<\delta$$ then $$|u|<\epsilon_f$$ and $$|v|<\epsilon_g$$ hence $$|w|<\epsilon$$.

If both $$f(a)$$ and $$g(a)$$ are zero, then $$\epsilon_f=\epsilon_g=\sqrt{\epsilon}$$ works.

Let $$M(x,y)=xy$$. $$xy-ab=((x-a)+a)((y-b)+b)-ab=(x-a)(y-b)+a(y-b)+b(y-a)$$ so that $$|xy-ab| \leq |x-a||y-b|+|b||x-a|+|a||y-b|=(|x-a|+|a|)(|y-b|+|b|)-|ab|$$ in case $$a \neq 0$$ and $$b \neq 0$$ we may divide both sides by $$|ab|$$ to get $$\frac{|M(x,y)-M(a,b)|}{|ab|} \leq \left(\frac{|x-a|}{|a|}+1\right)\left(\frac{|y-b|}{b}+1\right)-1$$ If $$\delta_x \leq |a| \left(\sqrt{1+\frac{\epsilon}{|ab|}}-1\right)$$ and $$\delta_y \leq |b| \left(\sqrt{1+\frac{\epsilon}{|ab|}}-1\right)$$ then for $$x$$ satisfying $$|x-a| \leq \delta_x$$ and $$y$$ satisfying $$|y-b| \leq \delta_y$$ wqe have $$|M(x,y)-M(a,b)|< \epsilon$$ If $$b=0$$ and $$a \neq 0$$ then we have $$xy=(x-a)y+ay$$ so $$|xy| \leq |x-a||y|+|a||y|=(|x-a|+|a|)|y|$$ by dividing both sides by $$|a|$$ we get $$\frac{|xy|}{|a|} \leq \left( \frac{|x-a|}{|a|} + 1 \right) |y|$$ For $$N$$ sufficiently large $$N\sqrt{\frac{\epsilon}{|a|}} > 1$$ choose $$\delta_x = |a|*\left( N\sqrt{\frac{\epsilon}{|a|}}-1 \right)$$ and $$\delta_y = \frac{1}{N} \sqrt{\frac{\epsilon}{|a|}}$$ then for $$|x-a| < \delta_x$$ and $$|y|<\delta_y$$ we have $$|xy| < \epsilon$$. If $$a=b=0$$ then for $$|x|<\sqrt{\epsilon}$$ and $$|y|<\sqrt{\epsilon}$$ we have $$|xy|<\epsilon$$. This proves that multiplication is continuous. Now if we assume that we can show that composition of continuous functions is continuous, then this shows the previous result as a special case. And such is the case,$$F$$ is continuous at $$b=G(a)$$ if and only if for every $$\epsilon>0$$ there exists $$\delta_F^{\epsilon}$$ such that for $$||y-b||<\delta_F^{\epsilon}$$ we have $$||F(y)-F(a)|| < \epsilon$$ and $$G$$ is continuous at $$a$$ if and only if for every $$\epsilon_2>0$$ there exists $$\delta_G^{\epsilon_2}>0$$ such that for $$||x-a||<\delta_G^{\epsilon_2}$$ we have $$||G(x)-G(a)||<\epsilon_2$$ choose $$\epsilon_2 = \delta_F^{\epsilon}$$ then for $$||x-a||<\delta_F^{\epsilon_2}$$ we have $$||F(x)-F(a)||<\epsilon_2=\delta_F^{\epsilon}$$ hence $$||G(F(x))-G(F(a))||<\epsilon$$. Hence composition of continuous functions is continuous. Also $$(x,y) \mapsto (f(x),g(y))$$ is continuous since given by componentwise continuous functions.

After reaching the step $$|f(x)g(x)-f(a)g(a)|\leq|f(x)-f(a)|*|g(x)|+|g(x)-g(a)||f(a)|$$ Given $$\epsilon$$ there exists $$\delta_g$$ such that for $$|x-a|<\delta_g$$ $$|g(x)-g(a)| < \frac{\epsilon}{2|f(a)|}$$ in case $$f(a) \neq 0$$. Now if $$g$$ is continuous near $$a$$ as well, then on the interval $$[a-\delta_g,a+\delta_g]$$ $$|g(x)|$$ assumes a maximum value say $$M=M_{\delta_g,a,g}$$. Similary there exists $$\delta_f$$ such that $$|f(x)-f(a)|<\frac{\epsilon}{2M}$$ whenever $$|x-a|<\delta_f$$ Now for $$|x-a|<\min(\delta_f,\delta_g)$$ the argument works. However this has the added assumption that functions be not only pointwise continuous but continuous on an interval around that point. But we can remove that $$|g(x)| \leq |g(x)-g(a)|+|g(a)| < \delta_g +|g(a)|$$ so choose $$M$$ above instead as $$M=\delta_g+|g(a)|$$. If $$f(a)=0$$ and $$g(a)\neq0$$ then $$|f(x)g(x)|=|f(x)*(g(x)-g(a))+f(x)*g(a)|\leq |f(x)|*|g(x)-g(a)| +|f(x)||g(a)|$$ for every $$\epsilon$$ there exists $$\delta_f$$ such that $$|x-a|<\delta_f$$ implies $$|f(x)|<\frac{\epsilon}{2|g(a)|}$$ and there exists $$\delta_g$$ so that $$|x-a|<\delta_g$$ $$|g(x)-g(a)|<|g(a)|$$. Now take $$\delta=\min(\delta_f,\delta_g)$$ If both $$f(a)=g(a)=0$$ then for every $$\epsilon$$ there exists $$\delta_f$$ so that for $$|f(x)| < \sqrt{\epsilon}$$ for $$|x-a|<\delta_f$$ and there exists $$\delta_g$$ so that for $$|g(x)|<\sqrt{\epsilon}$$.

In fact it is easiest if $$f(a)=g(a)=0$$. This is not the general case, but if $$f$$ is continuous at $$a$$ then $$s(x)=f(x)-f(a)$$ is also continuous at $$a$$. Here is why. for every $$\epsilon$$ there exists $$\delta_f^{\epsilon}$$ such that for $$|x-a|<\delta_f$$ we have $$|f(x)-f(a)|<\epsilon$$. But this is precisely the condition that $$|s(x)|=|s(x)-s(a)|=|f(x)-f(a)|<\epsilon$$. Similarly $$t(x)=g(x)-g(a)$$ is also continuous at $$a$$.

Now $$s(x)*t(x)$$ is continuous. Given $$\epsilon>0$$ there exists $$\delta_g^{\epsilon}$$ such that for $$|x-a|<\delta_g$$ we have $$|g(x)-g(a)|<\epsilon$$. Now for $$\delta=\min( \delta_f^{\sqrt{\epsilon}},\delta_f^{\sqrt{\epsilon}} )$$ when $$|x-a|<\delta$$ we have $$|s(x)*t(x)|<\epsilon$$. On the other hand, constant times a function is also continuous. so $$f(x)*g(x)=s(x)*t(x)+f(a)*g(x)+f(x)*g(a)-f(a)*g(a)$$ Since this is the sum of continuous functions it is continuous.