How to properly write a proof (in real-analysis)

Question

I'm working on how to write my proofs the way a "professional" mathematician does. From what I've seen, there are two very large differences between that and "school proofs":

1. Professional proofs omit anything that can be derived mechanically. You are not expected to "show your work," like in school.

2. Professional proofs explicitly and formally state things instead of appealing to arguments to intuition, even if the intuition might be convincing.

Both of these stem from the fact that the goal of a school proof is to show you know how to do it (hence, show your work), whereas, the goal of a professional proof is to provide proof to a peer.

However, I am having trouble doing this. For example, take this problem (from Berkeley Math 104 https://math.berkeley.edu/~gangliu/hw/final.pdf ):

If $f(x)$ and $g(x)$ are uniformly continuous and bounded, show that $u(x) = f(x)*g(x)$ is uniformly continuous.

The given solution ( https://math.berkeley.edu/~gangliu/hw/samp.pdf ) is "The proof follows if we use the definition of uniform continuity(details skipped)." That's not very satisfying.

Here is my proof:

For any $x$ and $x'$, define $\alpha = f(x) - f(x')$ and $\beta = g(x) - g(x')$. By definition of uniform continuity, for any $x$ and $x'$, $|x - x'| < \delta_f(\epsilon) \implies |\alpha| < \epsilon$, and $|x - x'| < \delta_g(\epsilon) \implies |\beta| < \epsilon$, with $\delta_f$ and $\delta_g$ functions of $\epsilon$ only. Write $\mu$ for $|u(x) - u(x')| = |\alpha g(x') + \beta f(x') + \alpha \beta|$. It suffices to show that there exists a $\delta_u$ as a function only of $\epsilon_u$, such that if $|x - x'| < \delta_u$, then $\mu < \epsilon_u$.

By the triangle equality, $\mu$ is no greater than the sum of the absolute value of it's terms. Since $f$ and $g$ are bounded, each term can be brought arbitrarily close to zero if $\alpha$ and $\beta$ are brought arbitrarily close to zero. As above, since $f$ and $g$ are uniformly continuous, for any non-zero value $k$, we can select a $\delta_f$ and $\delta_g$ within which $\alpha$ and $\beta$ will be smaller than $k$; hence, we can select a $\delta_u$ such that $\delta_u < \delta_f$ and $\delta_u < \delta_g$ which brings $mu$ arbitrarily close to zero. QED.

While I believe that my proof is correct, it's quite a mouthful, and hardly very clear. What is a good, professional, way to write this proof?

Note: I understand that there may be other, better, or shorter proofs. My question doesn't concern that. It's about how to write the proof, not what the best proof is.

Answer 1

After writing long comments I thought it is better to put all that stuff into a proper answer.

In general a proof has two aspects: rigor and formalism. Rigor is the part that deals with the essential/key ideas of the proof and is based on already agreed axioms and established theorems and laws of logical deduction. Formalism on the other hand is about using the mathematical symbols properly to make the proof unambiguous and precise. You can think of "formalism" as the "legalese of mathematics".

While writing a proof one can sacrifice formalism perhaps to make the proof more accessible to a wider audience, but you can't sacrifice rigor. A "non-rigorous proof" is a contradiction in terms.

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The proof given in your question is correct and the only improvement I can think of is to make it shorter in length by writing $$|u(x) - u(x')| = |\alpha g(x') + \beta f(x') + \alpha\beta| \leq |\alpha||g(x')| + |\beta||f(x')| + |\alpha||\beta|$$ and then saying that "since $\alpha, \beta$ can be made arbitrarily small and $f, g$ are bounded therefore each term on the right of the equation can be made arbitrarily small". Thus $|u(x) - u(x')| $ can be made arbitrary small by choosing $|x - x'|$ sufficiently small.

The above avoids the details of $\epsilon, \delta$ calculation which convey the formal meaning of phrases like "arbitrarily small" and "sufficiently small". For a person experienced in analysis proofs small calculations of $\epsilon, \delta$ (like in the current question) can be avoided because they are obvious. But if these calculations are not obvious then they should be expressed clearly using the proper symbols.

Obvious and small steps in a long proof can (and perhaps should) be avoided so that the reader gets to the core of the proof without getting bogged down into details (which are obvious anyway).

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Avoiding such steps is not hand-waving. Hand-waving always involves leaving out some essential idea of the proof and it sacrifices rigor. Hand-waving aims to pass an incomplete proof (or sometimes just garbage) for a legitimate proof.

Here is one example of hand-waving I just constructed:

Rolle's Theorem: If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$ and $f(a) = f(b)$ then there is a $c \in (a, b)$ for which $f'(c) = 0$.

Proof based on hand-waving: If $f$ is constant then we are done. So let's suppose $f$ is not constant and let's take the case when $f$ takes values larger than $f(a) = f(b)$ in the interval $[a, b]$. Then as we we move from $x = a$ to $x = b$ then $f(x)$ takes values larger than $f(a)$ and again falls back to $f(b)$. Hence $f$ increases and then decreases and thus there must be some point where $f$ changes its nature from increasing to decreasing. And at this point $f$ can neither increase nor decrease and hence $f'$ vanishes at that point.

The hand-waving is not in the informal language, but in leaving out the fact there must exist a point where $f$ switches from increasing to decreasing. This is the point where $f$ attains a maximum. Such a point is guaranteed because of the fact that $f$ is continuous on $[a, b]$. Notice that the proof in the previous paragraph never used the continuity of $f$ explicitly. Instead it chose to rely on the gullible nature of a naive reader who might think that continuous functions behave in this manner. The property of attaining a maximum value is a non-trivial property of continuous function which must be expressed explicitly. If this is included in the proof given above I would consider it as correct even though some extra formalism about increasing / decreasing might have been desirable.

Answer 2

Reflecting, I think the writing starts clear enough, but then trips over itself trying to show that something should be possible, instead of just constructing it. How about:

Write $b = max(|f(x)|, |g(x)|)$. If $b = 0$, $u(x) = 0$, and the proof is trivial. Otherwise, for any $x$ and $x'$, write $\alpha = |f(x) - f(x')|$ and $\beta = |g(x) - g(x')|$. Note that $|u(x) - u(x')| = |\alpha g(x') + \beta f(x') + \alpha \beta| \leq |\alpha g(x')| + |\beta f(x')| + |\alpha \beta| \leq |\alpha b| + |\beta a| + |\alpha \beta|$. Call this last quantity $\mu$.

Since $\mu$ is the sum of terms which can be brought arbitrarily close to zero, $\mu$ can likewise be brought arbitrarily close to zero, as will show.

For any $\epsilon$, let $\epsilon' = min(\sqrt{\epsilon/3}, \epsilon/3b)$. Since $f$ and $g$ are uniformly continuous, there exists a $\delta$ such that if $|x - x'| < \delta$, $\alpha < \epsilon'$ and $\beta < \epsilon'$, so that $\mu < \epsilon$. QED.

The interesting thing here is that the proof is written in the reverse order which I came up with it. For example, defining $b$ as the max bound was the last step in my mind, but the first step in the writing.