Let $f$ be a $C^1$ functional on a Hilbert space $X$, and $Y$ a closed subspace of $X$. Suppose the restriction of $f$ on $Y$ has a critical point $x_0 \in Y$.
Q: Is $x_0$ a critical point of $f$?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Of course not. Take e.g. $f$ a non-zero linear functional and Y=ker(f). $\endgroup$– Pietro MajerCommented Feb 3, 2012 at 12:55
-
$\begingroup$ (or also, whatever X and f, and Y=(0)). $\endgroup$– Pietro MajerCommented Feb 3, 2012 at 13:20
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Especially in Calculus of Variations and Mechanics, a submanifold $Y$ of a manifold $X$ is usually called "a natural constraint" for a functional $f$ on $X$, if the special circumstance that you are considering does happen: constrained critical points of $f$ on $Y$ are free critical points: $\operatorname{crit} (f_{|Y})\subset \operatorname{crit}(f)$.
An important situation when this is true arises when $Y$ is defined as fixed point set of a group of symmetries of $f$. The corresponding principle, claiming that $Y$ is a natural constraint, is known since old times as a popular proverb. In the late 70' Richard Palais has made this into a deep and beautiful theorem, stating the right hypotheses for its validity and also providing counter-examples to the naive claim of the principle (all popular proverbs are 95% true but never 100% true!). [ R. Palais, The Principle of Symmetric Criticality, Comm. Math. Phys., Vol. 69, Number 1 (1979), 19-30 ]
answered Feb 3, 2012 at 13:33