Let X be a Banach space and let T be a bounded linear operator on X. We know that if X is reflexive and T is compact then there exists x in the unit sphere of X such that T attains its norm at x. Can we impose any condition on X or on T such that x is the unique such point?
Guliyev is right. Even in finite dimensional case, the point x with its unitary multiples need not be unique. e.g. the norm max(|x_1|, |x_2|) in R².
There are some results beeing improvements of the Krein-Rutman Theorem. If you have an strongly positive and compact operator T on a Banach space with a cone then its spectral radius r(T) is a simple eigenvalue and other elemets of the spectrum has smaller modulus. You can find it in Yihong Du, Order structure and topological methods in PDEs , in introduction.
Bogdan Przeradzki
Since the answer is no, the question could be asked as follows: is there an x such that whenever the norm of T is attained at some y, then y is a multiple rx of x, with |r|=1? This is a kind of uniqueness.
In general, it is a difficult problem to identify the points at which an operator attains norm, if it attains norm at all! Having said that, there are certain estimations available in l_p^2 spaces over the real field, where p \neq 1, 2, \infty. It says that if an operator attains norm at more than 2(8p-5) number of points, then it must be a scalar multiple of an isometry.
For p=2, norm attaining point is unique (upto multiplication by \pm 1) if and only if the concerned operator is a smooth point in the operator space. In general, for a SMOOTH operator between Banach spaces, norm attaining point is unique (upto multiplication by \pm 1) if the operator attains norm.
Another easy sufficient condition is to assume that X is reflexive, strictly convex and T is rank 1. In this case it is trivial to observe that M_T ={\pm x_0}, for some x_0 in S_X, the so called unit sphere of X.
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