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Condition (C). The closure of any nonempty subset S of H on which f is bounded but on which $\|\nabla f\|$is not bounded away from zero, contains a critical point of f. How to see the meaning of " $\|\nabla f\|$is not bounded away from zero"?

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To get acquainted with this notion, you may start from the following exercise: prove that a $C^1$ functional on a Banach space, bounded below and satisfying the Palais-Smale condition (C), is coercive (i.e., sublevel sets are bounded). – Pietro Majer May 14 '12 at 5:07
Thank you for your answer. But, I cann't still understand the meaning of this sentence of $\|\nabla f\|$ is not bounded away from zero". – unknown (yahoo) 0 secs ago – economic Jun 14 '12 at 2:20

The condition forces things that look like critical points, to be critical points. Through various techniques, e.g. minimax methods, linking methods, morse theory, one obtains sequences that seem to converge to critical points. That is sequences $x_i$ that satisfy $||\nabla f(x_i)||\rightarrow 0$. and $||f(x_i)||$ is bounded. Let us call these Palais-Smale sequences. If a function/functional $f$ satisfies: all Palais-Smale sequences for $f$ have converging subsequences, we say $f$ satisfies the Palais-Smale condition.

In general Palais-Smale sequences do not need to have a subsequence which converge to a limit. A very elementary and standard example on $\mathbb{R}$ is the following. Let $f(x)=\arctan(x)$, and the sequence $x_i=i$. It is easy to verify that the sequence is Palais-Smale, but does not have a converging subsequence.

In this example we could get away with another property: properness (the preimage of every compact is compact). However, if $f:M\rightarrow\mathbb{R}$ is defined on a non-locally compact space (e.g. Banach/Hilbert spaces/manifolds), it can never be a proper map. The Palais-Smale condition is a condition which is satisfied by many interesting functionals, and ensures the convergence issues one would like.

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Thank you for your answer. But, I cann't still understand the meaning of this sentence of $\|\nabla f\|$ is not bounded away from zero". Is $\|\nabla f\|>0$ no bounded below, and verge to 0? – economic Jun 14 '12 at 2:26
There is no constant c>0 such that $||\nabla f||>c$ on $S$. – Thomas Rot Jun 14 '12 at 5:39
Thus if this condition does not hold, there is a sequence such that $|\grad f(x_i)|$->0 in S. The condition forces this sequence to have a limit, and the limit must be a critical point. – Thomas Rot Nov 23 '12 at 9:43

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