Bag of tricks in Advanced Calculus/ Real Analysis/Complex Analysis

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\epsilon > 0:x \leq y + \epsilon$.

Answer 4

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.