Why using liminf instead of limsup?

Question

In Chapter 8: Calculus of variations of Evan's Partial Differential Equations, Evan writes as follows:

I am wondering about the last paragraph where he says that

knowing $I[u] \leq \liminf_{j\to\infty} I[u_{k_j}]$ is sufficient to claim that $u$ is a minimizer.

Why do we use $\liminf $ here instead of $\limsup$?

Woulnd't knowing $I[u] \leq \limsup_{j\to\infty} I[u_{k_j}]$ is sufficient as well to conclude that $u$ is a minimizer since we know $lim_{k\to\infty} I[u_k] = m = \liminf_{j\to\infty} I[u_{k_j}] = \limsup_{j\to\infty} I[u_{k_j}]$ ?

If so, since $\liminf_{j\to\infty} I[u_{k_j}] \leq \limsup_{j\to\infty} I[u_{k_j}]$, wouldn't it be "easier" to show

$$I[u] \leq \limsup_{j\to\infty} I[u_{k_j}]\text{ ?}$$

than it is to show

$$I[u] \leq \liminf_{j\to\infty} I[u_{k_j}]\text{ ?}$$

Answer 1

wouldn't it be "easier" to show

A weaker statement is not always easier to prove. See Examples where it is easier to prove more than less.

You are right: to derive that $u$ is a minimizer it would suffice to show that $$I[u] \leq \limsup_{j\to\infty} I[u_{k_j}] \tag{1}$$ Why does Evans talk about the stronger property $$I[u] \leq \liminf_{j\to\infty} I[u_{k_j}] \tag{2}$$ instead? Because (2) is a well behaved property of functions (known as lower semicontinuity). For example:

• the supremum $\sup_\alpha u_\alpha$ of any family of lower semicontinuous functions is lower semicontinuous.
• $u$ is lower semicontinuous on a set $U$ if and only if the set $\{(x,z):z\ge u(x)\}$ is closed in $U\times \mathbb{R}$

Also, if (1) holds for every sequence converging to $u$, then (2) holds as well. Indeed, to prove (2) one only needs to pick a subsequence converging to $\liminf$, and apply (1) to that subsequence.