Where does the mass disappear in Fatou's Lemma?

I am looking for an intuitive explanation for Fatou's Lemma. I can follow the proof using Monotone convergence theorem, also understand the usual examples, but I do not know how to explain the inequality in words.

What is exactly that makes $\liminf_n E(f_n)-E(f) \ge 0?$ If I may ask, why does some sort of law of conservation not hold here?

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If you take $A$ and $B$ disjoint of positive measure, then in $\int \liminf f_n d\mu$, where $f_{2n}$ is the characteristic function of $A$ and $f_{2n+1}$ the characteristic function of $B$, we get $0$, since it jumps from $A$ to $B$ and from $B$ to $A$, but the $\liminf$ of the integral is $\min(\mu(A),\mu(B))$. – Davide Giraudo Mar 6 '12 at 17:38
I think this example is one where a sequence oscillates between two values and the lim inf is the minimum of the two. It is a bit specific, I think. – Bravo Mar 7 '12 at 4:06
Yes, it's only an example. Sam's answer is much more complete. – Davide Giraudo Mar 7 '12 at 13:45

You are right to consider mass escape. In fact in analysis, there are only a few ways for mass to escape or to lose compactness:

1) Take $f_n$ to be a bump that slides off to infinity

2) Take $f_n$ to be successive approximations of a point mass measure, say $f_n=\frac{1}{2} n1_{[-1/n,1/n]}$.

3) Take $f_n$ to be approximations of a wildly oscillating function, say sin(n*x)

The first example is what can happen when you are not on a compact set, and the last two can happen on compact sets. In the first two cases, $f_n\rightarrow 0$ almost surely, whereas the last function converges weakly to zero: $\int_{[0,1]} f(x)\sin(nx)dx\rightarrow 0$ for all $L^1[0,1]$ functions (Riemann-Lebesgue Lemma). Your interest is mostly in the first two cases since you want to be dealing with POSITIVE functions. So, whenever your sequence of functions $f_n$ doesn't satisfy the stronger conditions of the Monotone Convergence Theorem (that the sequence is INCREASING, and therefore mass cannot escape), usually the best you can say is what Fatou's lemma gives you.

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You and Sam have already given hints toward what I normally provide as a 'wordy' description of Fatou's Lemma, but not written it explicitly.

To answer the question, I suggest you use: "Energy can not be gained through relaxation".

I keep in mind the standard example where $$f_n(x) = \begin{cases}n,\text{ for }x\in[-1/(2n),1/(2n)]\\ 0,\text{ otherwise}, \end{cases}$$ where $x\in\mathbb{R}$ and we use standard Lebesgue measure on $\mathbb{R}$. Then the measure of each $f_n$ is 1 for every $n$, i.e. each element of the sequence has unit energy, but $f_n$ converges uniformly to the zero function (I say "$f$ relaxes to zero"), which has measure zero, and so zero energy.

The example proves the obvious statement that "Energy can be lost through relaxation". It also begs the question of whether or not one may gain energy through relaxation.

Fatou's Lemma gives the non-trivial answer: "Energy can not be gained through relaxation".

Caveat: It is not always appropriate to think of limits of functions as a 'relaxation'. Plenty of things can blow up in a limit (the derivative of a smooth approximation of a discontinuous function must blow up for example). Still, I find it to fit here very well.

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Thanks Glen. While I get the gist of what you say, could you explain what exactly is 'relaxation'? Is it just the equilibrium, like that of a spring that stops oscillating? If it is so and if the zero function is that equilibrium, do you envisage the entire process as a contraction of a spring to zero? – Bravo Mar 7 '12 at 15:26
@Shyam I (personally) visualise a sequence of functions as being time-slices from some abstract energy-minimising evolution equation. Relaxation is just a synonym for taking a limit of these time-slices, and is a reduction of energy ('Excitation' being the opposite). Relaxation doesn't always give an equilibrium. In the sense it is used here, it is just a synonym for taking a limit. – Glen Wheeler Mar 7 '12 at 15:39