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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.

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I like this question! A soft-question tag would probably be appropriate here, though. –  Dahn Jahn Jan 11 '13 at 1:27
    
and/or "big-list" tag: probably should be CW, as well. –  amWhy Jan 11 '13 at 1:30
    
@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. –  Jack Dawkins Jan 11 '13 at 1:32
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Tim Gowers has an entire wiki devoted to this sort of thing: tricki.org. –  kigen Jan 11 '13 at 1:40
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6 Answers 6

up vote 5 down vote accepted

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.

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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. –  Jack Dawkins Jan 11 '13 at 18:40
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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.

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Use the fact that $\log(ab) \ge \lambda \log(a) + (1 - \lambda) \log(b).$ choose teh right values for $\lambda$, $a$ and $b$. –  ncmathsadist Jan 11 '13 at 1:52
    
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. –  A Blumenthal Jan 12 '13 at 21:04
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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 in 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.

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

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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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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 $n\in\Bbb N$.
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