I am studying for an exam and I have been studying my butt off during the winter break for it. During the course of my study I have written down quite a number of tricks, which in my opinion were 'outrageous' :-). Meaning there was no way I would come up with that during an exam if I hadn't seen that before.

Couple of examples.

  1. Sometimes, when you want to prove something about $\max$, $\min$, you write ( I got this from Baby Rudin)

$$ \max(a,b)=\frac{a+b+ \vert a-b \vert} {2} $$ $$ \min (a,b)= \frac{a+b-|a-b|} {2} $$

  1. To prove Hölder's inequality (in its simplest case) You write $\int (f+tg)^2 \geq 0$ and since this stays positive you get that the discriminant of this must be negative, and magically you get your Hölder inequality.

  2. When you want to show something about distinct zeroes of complex functions you kind of eliminate the zeroes of f by dividing them with the appropriate Möbius transforms and you still get an analytic functions which has nice properties.

The value of these is that they can be used in other contexts to write neat proofs.

That's what I mean by "tricks". This might be difficult to answer, but what are some of the tricks you wise folks have up your sleeve when it comes to Advanced Calculus (Both single variable, multivariable) and complex Analysis.

Anything you have to share will be greatly appreciated. Thanks so much for all your help.

  • 1
    $\begingroup$ I like this question! A soft-question tag would probably be appropriate here, though. $\endgroup$
    – Dahn
    Commented Jan 11, 2013 at 1:27
  • 1
    $\begingroup$ and/or "big-list" tag: probably should be CW, as well. $\endgroup$
    – amWhy
    Commented Jan 11, 2013 at 1:30
  • $\begingroup$ @DahnJahn: I am not sure how to do either of those. I am new here. How do I go about putting them?. You can go ahead and make the required modifications if you want. $\endgroup$
    – Cousin
    Commented Jan 11, 2013 at 1:32
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    $\begingroup$ Tim Gowers has an entire wiki devoted to this sort of thing: tricki.org. $\endgroup$ Commented Jan 11, 2013 at 1:40
  • 1
    $\begingroup$ Could you please provide more details regarding the trick of Holder's inequality? $\endgroup$
    – FNH
    Commented Nov 30, 2016 at 16:57

7 Answers 7


Here are a couple of tricks and general plans of approach I know:

  • If $x\in\Bbb R$ and for all $\epsilon>0$ we have that $|x|\leq\epsilon$, then $x=0$. I think of this fact as being Real Analysis in a nutshell.
  • Never forget that if $A\subseteq\Bbb R$ is bounded above and $M=\sup A$, then for all $\epsilon>0$ there is a $y\in A$ such that $M< y+\epsilon$. Likewise if $A$ is bounded below and $m=\inf A$, then for all $\epsilon>0$ there is a $y\in A$ such that $y-\epsilon< m$. This is the most important tie between $\Bbb R$'s algebraic and ordering properties.
  • Never underestimate the binomial theorem even if you just want an inequality. As an example, look at theorem 3.20(c) in Rudin and how he uses it.
  • For all $x,y\in\Bbb R$ and any $\epsilon> 0$, we have the following inequality: $$|xy|\leq \frac{\epsilon\,x^2+\epsilon^{-1}\,y^2}{2}$$ This can be derived from the observiation that $(\epsilon\,|x|-|y|)^2\geq 0$. This inequality allows us to decide how much 'weight' we want to give to a particular term in a product. This can be used to show that the product of Riemann integrable functions is still Riemann integrable.
  • A simple inequality to remember is $$(a+b)^p\leq 2^p(a^p+b^p)$$ for $a,b,p\geq 0$. This can be derived from the even simpler inequality $(a+b)\leq 2\max(a,b)$, again for positive values. This inequality can be used to show that the $L^p$ spaces are vector spaces.
  • The Weierstrass M-test is the first friend you call when dealing with series of functions.
  • Ask yourself whether the problem you're working on can be generalized to topology first. Think about compactness and connectedness and the abstract theorems about them you already know.
  • Perhaps the greatest topological property that $\Bbb R$ has is second-countability. This means that $\Bbb R$ is hereditarily-separable, sequential, Frechet-Urysohn, and c.c.c. This property of $\Bbb R$ allows us to consider sequences and sequential continuity in place of neighborhoods and continuity. As an adage, if you are working with $\epsilon$, wonder to yourself if you can instead work with $1/n$ with $n\in\Bbb N$.
  • $\begingroup$ Thanks for a detailed and a nice answer. $\endgroup$
    – Cousin
    Commented Aug 17, 2014 at 4:46

One `trick' that is used a lot in my analysis courses is: instead of showing that $x \leq y$ directly, it is usually a lot easier to show that, for all $\epsilon > 0:x \leq y + \epsilon$.


I'm not sure if you are requiring that a trick be something overly hard/creative or just something more along the lines of "Ahhh... I might not have thought of that, but now that I've seen it, I'd be able to do that again!", especially if it appears again and again.

If you mean the latter, keep in mind the ol' "add-and-subtract" or "$\varepsilon/3$ trick" where you insert a new term(s) that adds and then subtracts off some useful quantity, usually is followed by an appeal to the triangle inequality (or something similar) and some known estimates. A classic example is in proving that the uniform limit of continuous functions is continuous, where we use this to manufacture the terms leading to the $\varepsilon/3$'s.

It's not a complicated technique but certainly a recurring one in analysis.


The reverse triangle inequality $$ |z - w| \geq ||z| - |w|| $$ - note the iterated absolute value signs on the right - is very useful to prove an integral of the form $\int_\gamma (f(z)/g(z))\,dz$ is small in the course of evaluating a real integral via the residue theorem. By the triangle inequality $|\int_\gamma f(z)/g(z)\,dz| \leq \int_\gamma |f(z)/g(z)|\,dz$, but to get an upper bound on $|f(z)/g(z)| = |f(z)|/|g(z)|$ we need a way to form a worthwhile lower bound on $|g(z)|$. If $g(z) = u(z) - v(z)$ in some natural way, then $$ \left|\int_{\gamma}\frac{f(z)}{u(z) - v(z)}\,dz\right| \leq \int_\gamma \left|\frac{f(z)}{u(z)-v(z)}\right| \leq \int_{\gamma} \frac{|f(z)|}{||u(z)| - |v(z)||}\,dz, $$ and now the underlying geometry of the situation may help us understand $|u(z)|$ and $|v(z)|$ separately on the contour $\gamma$ in order to make further progress.

When I was first learning to use the residue theorem in calculations of real integrals, I was quite impressed when I first saw this inequality in action, and then I found myself using that idea all the time on such problems to prove some contour integral was small.

The inequality itself is easy to derive using the add-and-subtract idea mentioned by JohnD: $|z| = |z-w+w| \leq |z-w| + |w|$, so $|z-w| \geq |z| - |w|$. Swapping the roles of $z$ and $w$ then gives $|z-w| \geq |w| - |z|$. One of $|z| - |w|$ or $|w| - |z|$ is $||z| - |w||$ (the other is $\leq 0$), and the reverse triangle inequality falls out.

  • $\begingroup$ Yes this is a very good estimate that is almost always used in computing such integrals. Another useful trick in the so called Jordan's lemma, which helps show that terms like $\int_0^{2\pi} e^{-aRsin t} dt$ goes to zero as $R$ goes to infinity. $\endgroup$
    – Cousin
    Commented Jan 11, 2013 at 18:40

I would say the main recurrent goals in analysis is to prove that two things are either sufficiently close together or arbitrarily close together (and thus equal).

These rather vague goals, in my opinion, feed directly into the appeal and infamy of analysis. For there are certainly uncountably many ways to show that two values of interest are close together and the common recurring tactic here is to do so by proxy: this is close to that, that is close to yonder, yonder is close to over there.... therefore this is close to over there. Knowing which intermediary points to compare things to is partly craft but also partly art.

Any concrete recommendations (including my past ones in my other answer) I would call part of the "craft" of analysis. You learn tricks of astonishing mastery by other mathematicians and you mimic their approach when you think you're in a similar situation. However the "art" of analysis is puttying your own spin on these tricks first by applying some sort of method to the madness---find some way of grouping these tricks that makes sense to you---and then wait for the method to inspire new madness.

Some of the methods that I have collected under analysis include:

  • Collect inequalities.
  • Collect algebraic identities.
  • Collect limits (even slowly converging ones) in their various forms.
  • Exploit density, compactness, and connectedness.
  • Attempt to seesaw the quantities you're working with.
  • Give yourself some buffer room.
  • Introduce a variable, even if it makes your problem seemingly more difficult.
  • Take the Herglotz point-of-view.
  • Be optimistic.

each of which I should probably describe by some example. Of course, the underlying form of just about all analytical arguments include inequalities. Thus, it is helpful to have several different ones at your fingertips. Particularly useful ones are inequalities that connect algebraic functions with transcendental functions such as $$1+x\leq e^x\,,\quad \frac{2}{\pi}|x|\leq |\sin x|\leq |x|$$ or inequalities that attach several different "levels of arithmetic" with each other in different orders such as Cauchy's Inequality which trades sums of products for products of powers of sums of powers. I would say that almost all of these inequalities can be derived from convexity and Jensen's inequality.

However, it is also frequently useful to know certain algebraic identities. Many interesting assertions have clever proofs that involve rewriting the quantities in question by some algebraic manipulation where suddenly everything is trivial with hindsight. These proofs were concocted with algebraic insight. Knowing identities with sums of squares, Vieta's Formulae, the Binomial Theorem, special factorizations, factoring polynomials of low degree, exploiting telescoping, and trigonometric identities (as a couple examples) can all feed into making a very sleek and cheeky proof. If you happen upon a novel identity. Keep it somewhere.

Sometimes, you only care about what happens in the limit than at a particular point in time. In this case, having several limits on hand are extremely useful. The more routine limits arise from differentiation; however, you can also find nifty limits by realizing things as Riemann sums. More arcane limits, especially slow ones for some reason, are useful too. You can demonstrate that $\Gamma(1/2)=\sqrt{\pi}$ (and thus calculate the Gaussian integral) with the Wallis product and $\Gamma$'s Euler product form. The various sequences that converge to $e$ also notoriously spring up like weeds.

The next point addresses topological arguments. There are several nice theorems that only occur on compact spaces or connected ones. Keep them. Density however creeps up quite a bit as well. If you want to show two functions or functionals are equal, you need only do so on a dense subset, which can dramatically simplify the argument you have to do.

Although a lot of analytic approximations are through a chain of proxies, some arguments are done by "seesawing". That is, you reallocate some weight from some term over to another term that wouldn't mind having the extra weight. The "adding-and-subtracting" trick can be seen as falling in here. But I would also toss the Peter-Paul inequality in my last answer and Young's Inequality here.

"Giving yourself some buffer room" roughly translates to attempting to prove a stronger result which will give you your original result upon taking a limit. For example, if you want to show $$\sum_{k=1}^\infty\frac{1}{k^2}\leq 2$$ you should instead show $$\sum_{k=1}^n\frac{1}{k^2}\leq 2-\frac{C}{n}$$ for some constant $C$. Rephrasing the question like this suddenly opens up the door for induction. And the argument will take you there.

Introducing a variable, in my opinion, is perhaps the dirtiest trick I can think of in elementary analysis. For example, to calculate the improper integral $$\int_0^\infty \frac{\sin x}{x}\,dx$$ you should instead attempt to calculate $$\int_0^\infty\frac{\sin x}{x}e^{-tx}\,dx$$ for arbitrary $t$ where $t>0$. Or to calculate the sum below $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}$$ you should instead calculate the sum $$\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}$$ for arbitrary $x$ with $|x|<1$. The user Jack D'Aurizio on this site is a master of this technique and also a master of rewriting things as integrals. View his answers to get a feel for this.

Herglotz is mildly famous for giving a fairly elementary and easy-to-follow proof of the following identity: $$\pi\cot\pi x=\lim_{N\rightarrow\infty}\sum_{k=-N}^N\frac{1}{x+k}\,.$$ Everyone needs to read that proof and digest it for themselves. But the essential idea behind the proof is to demonstrate that the two functions on the left and right have enough properties in common to prove that their difference is zero. This idea also swallows up the idea of proving that two differentiable functions are equal if their derivatives are equal and they agree at a point. A very similar proof to Herglotz's also proves the $\Gamma$ reflection formula with sine.

"Be optimistic" is meant to encapsulate the idea that in order to construct something, it is sometimes helpful to just assume that it exists and deduce necessary properties that hint at its construction. Bohr and Mollerup gave a uniqueness proof of the $\Gamma$ function that follow more along this motif rather than Herglotz's motif. A similar idea can be used to give a rather strange-looking definition of sine and cosine if one can only talk in the language of elementary analysis.

Of course, there are several other ideas that I could write. But this answer is already more like a blog post than a StackExchange answer.

  • $\begingroup$ Nice answer and your remark about Jack D'Aurizio is quite true. +1 $\endgroup$
    – Paramanand Singh
    Commented May 3, 2018 at 3:07
  • $\begingroup$ @ParamanandSingh a compliment from you is flattery. Thank you! Always impressed by your answers as well. $\endgroup$
    – user123641
    Commented May 3, 2018 at 13:21

Now hear this. Suppose that $1/p + 1/q = 1$. Then if you exploit the convexity of the log function you can show that for $x, y \ge 0$ $$xy \le {x^p\over p} + {x^q\over q}.$$ This is pivotal in proving Hölder's inequality.

  • $\begingroup$ Use the fact that $\log(ab) \ge \lambda \log(a) + (1 - \lambda) \log(b).$ choose teh right values for $\lambda$, $a$ and $b$. $\endgroup$ Commented Jan 11, 2013 at 1:52
  • $\begingroup$ Anything related to the convexity/concavity of functions is useful. For instance, the concavity of $x \mapsto \log x$ gives the above useful inequality in the proof of Holder. Another useful one: $x \mapsto e^{-x}$ is used, for instance, in the proof of the second (hard) part of the Borel Cantelli Lemma, where the inequality $1-x \leq e^{-x}$ is used. The general principle is that a C^1 convex function dominates its tangents and is dominated by its chords; a concave function is dominated by its tangents and dominates its chords. $\endgroup$ Commented Jan 12, 2013 at 21:04

The sandwih trick: if $a_n\le b_n\le c_n$ and $\lim a_n=\lim c_n=L$ then $\lim b_n=L$.


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