What is the sense of the equality sign =? Sometimes in real analysis, we write an equality and specify that is true in the sense of the $L^2$ norm.
My question is, when in mathematics we write '=' and don't specify anything, what we really mean?
Do we mean uniform convergence, more than that, or less than that?
 A: When mathematicians write the equality sign $=$, they officially mean just equality, i.e. $x = y$ means $x$ and $y$ are the same thing.
It is sometimes convenient to allow $=$ to have a slightly different meaning, but most frequently this boils down to properly understanding what is on either side of the equality. For instance, if you want write $f = g$ for equality in $L^2$, what you really mean is that you treat $f$ and $g$ as elements of $L^2$, and you are comparing these. Hence, the problem boils down to the question: what are the objects you are talking about? (which is not quite trivial, since the "same" function may belong to a lot of various function spaces, for example). Note that a purist would probably specify by adding after $f = g$ something like "in $L^2$" or "almost everywhere".
If you have a limit or a series (which is a limit in disguise) on either side, the real question is: how are you taking the limit? To make sense of a statement like "$f = \sum_n a_n x^n$" or some other series, you need to figure out in what space you are taking the sum. Once you know this, he equality is the equality in that space. 
A: Building on Jesse Madnick's excellent answer, one way of interpreting

"$f = g$ in the sense of the $L^2$ norm"

is that what it really means is

"$\langle f \rangle = \langle g \rangle$, where $\langle f \rangle$ and $\langle g \rangle$ denote that $L^2$ equivalence classes of $f$ and $g$ respectively"

but we're being lazy in our notation and using just the plain function symbols $f$ and $g$ to stand for their equivalence classes, in order to avoid an excessive profusion of brackets.
Or, alternatively, we could interpret $f$ and $g$ as being the the equivalence classes, and say that we're being lazy by using them elsewhere as functions without explicitly picking a representative function from each class.
Either way, this habit of using the same symbol to denote both an equivalence class and a representative member of it is common in many areas of mathematics, since it makes the notation much less cluttered.  It does, however, mean that we sometimes need to explicitly clarify in which sense we are using a particular symbol.
A: The relationship $f\sim g$ defined as $\|f-g\|=0$ is called an equivalence relationship.
In general, if $X$ is a set, an equivalence relation on $X$ is a subset of $R\subseteq X\times X$ which has the properties:


*

*Reflexivity: $\forall x\in X:(x,x)\in R$

*Symmetry: $\forall x,y\in X: (x,y)\in R\implies (y,x)\in R$

*Transitivity: $\forall x,y,z\in X: (x,y)\in R \text{ and }(y,z)\in R\implies (x,y)\in R$


If we write $(x,y)\in R$ as $x\sim_R y$ then this means 


*

*$x\sim_R x$

*$x\sim_R y \implies y\sim_R x$

*$x\sim_R y \text{ and }y\sim_R z\implies x\sim_R z$


Now, given an equivalence relation on $X$, we can partition $X$ into something called equivalence classes, and we can construct a set $X/\sim_R$ and a map $\pi: X\to X/\sim_R$ such that $\pi(x)=\pi(y)$ if and only if $x\sim_R y$. We sometimes write $\pi(x)=\left<x\right>_{\sim_R}$ or just $\left<x\right>$ if the equivalence relation is known.
So, in your case, we start with $L^2$, and define an equivalence relation $\sim$ as above, and then construct $L^2/\sim$.
It is common to abuse notation, and instead of writing for $x,y\in L^2$ that either $x\sim y$ or that $\left<x\right>=\left<y\right>$, we just write $x=y$. This is lazy notation, but in cases like $L^2/\sim$, it is not a big deal.
You often want the equivalence relation to be nice. For example, in this case of $L^2$ and $\sim$, if $f_1\sim g_1$ and $f_2\sim g_2$ then $\|f_1-f_2\|=\|g_1-g_2\|$. Nice features of the equivalence relationship means that we don't have to take much care which $f$ we choose from an equivalence class in our proofs - if $\left<f\right>=\left<g\right>$ then the proof works the same using $g$ instead of $f$. However, we do have to prove that the equivalence relation is "nice" in this way.
For another example of the "niceness" of $\sim$ on $L^2$: If $f_1\sim g_1$ and $f_2\sim g_2$, then $f_1+f_2\sim g_1+g_2$. So we can define $+$ in $L^2/\sim$.
A: I think I see where the confusion is.  Let's clarify: generally speaking, the equality sign $=$ pretty much always means the same thing (two elements are the same in a given space).  It's the issue of convergence that changes.  This is best seen by two examples:
Example 1: Suppose I say that $\lim_{x \to 2} f(x) = 7$.  Now, one might say that this equation means that:

For all $\epsilon > 0$, there exists $\delta > 0$ such that $|x - 2| < \delta$ implies $|f(x) - 7| < \epsilon$.

So, one could mistakenly think, "Wow, that $=$ sign is hiding a lot of information.  That's a completely different kind of equality."  But it's not the $=$ sign that's doing it!  It's the limit itself.  In other words, we could have written it this way:
$\lim_{x \to 2} f(x)$ is the number such that

For all $\epsilon > 0$, there exists $\delta > 0$ such that $|x - 2| < \delta$ implies $|f(x) - \lim_{x\to 2} f(x)| < \epsilon$,

and this number, $\lim_{x \to 2} f(x)$, is (i.e. equals) 7.
Example 2: In a similar way, suppose $f_n$ are $L^2$ functions, and I say that $\sum_1^\infty f_k = g$ in $L^2$.  Again, this is best phrased like this:
$\sum_1^\infty f_k$ is the function such that

For all $\epsilon > 0$, there exists $N$ such that $n \geq N$ implies $\Vert \sum_1^n f_k - \sum_1^\infty f_k \Vert_2 < \epsilon$,

and also $g$ is that same function $\sum_1^\infty f_k$ (almost everywhere).
Summary: The equals sign $=$ is the same thing you've always known and loved (except that it's taken relative to a given space).  What you have to understand is what expressions like $\lim f(x)$ and $\sum f_n$ really mean.
