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

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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

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

Michael Hardy
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mononono
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  • I didn't read everything, because I think you may be getting a bit lost in details. Being below the limsup of a sequence (resp. above the liminf of a sequence) is an extremely weak condition. I can take a nice sequence $u_k$ that is tending towards a minimizer and interlace it with a sequence $v_k$ where $I[v_k]$ blows up. If the resulting sequence is $w_k$, then any $w$ would satisfy $I[w] \leq \limsup I[w_k]=\infty$, even though $w_k$ certainly contains a subsequence with the desired property. – Ian Apr 23 '15 at 23:36
  • @Ian $w_k$ would not be a minizming sequence though – mononono Apr 24 '15 at 01:05

1 Answers1

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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.

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