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

Solution 1:

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 $n\in\Bbb N$.

Solution 2:

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