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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.
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    $\begingroup$ 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 $s_{n}t_{n}-st=(s_n-s)(t_n-t)+s(t_n-t)+t(s_n-s)$. $\endgroup$
    – Fei Li
    Aug 24 '16 at 4:57
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    $\begingroup$ I think using the sequences definition of continuity is the neatest way of solving this question as well. $\endgroup$
    – gowrath
    Aug 24 '16 at 5:11
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    $\begingroup$ 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. $\endgroup$
    – fleablood
    Aug 24 '16 at 7:40
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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.

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    $\begingroup$ This is a brilliant proof. Anyone new to epsilon-delta proofs can test their conceptual understanding with this answer $\endgroup$
    – router
    Oct 15 '19 at 10:05
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    $\begingroup$ 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) $|f(x)| < \epsilon + |f(a)|$. $\endgroup$ Dec 30 '20 at 21:43
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    $\begingroup$ nice It should be saved $\endgroup$ Jun 18 '21 at 22:47
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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 $\epsilon$: $$\epsilon_f<\min\left(\frac{\epsilon_{f\cdot g}}{3|g(a)|},\sqrt{\frac{\epsilon_{f\cdot g}}3}\right),\\ \epsilon_g<\min\left(\frac{\epsilon_{f\cdot g}}{3|f(a)|},\sqrt{\frac{\epsilon_{f\cdot g}}3}\right).$$

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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

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

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