Proof verification: the product of two continuous functions is continuous

Solution 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.

Solution 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 $\delta_{f\cdot g}$.

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