Proof verification: the product of two continuous functions is continuous

Question

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.

Answer 1

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.

Answer 2

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 $\delta_{f\cdot g}$.

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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 $\epsilon$: $$\epsilon_f<\min\left(\frac{\epsilon_{f\cdot g}}{3|g(a)|},\sqrt{\frac{\epsilon_{f\cdot g}}3}\right),\\ \epsilon_g<\min\left(\frac{\epsilon_{f\cdot g}}{3|f(a)|},\sqrt{\frac{\epsilon_{f\cdot g}}3}\right).$$

Answer 3

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.

Answer 4

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.

Answer 5

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 $|g(x)|<\sqrt{\epsilon}$.

Answer 6

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.

Answer 7

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

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