From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: strict convex norm?
Date: 24 May 2000 15:07:58 -0400
Newsgroups: sci.math
Summary: [missing]

In article <8gh26u$joi$1@nnrp1.deja.com>,
 <num_anal_student@my-deja.com> wrote:
:Someone claimed that the frobenius norm is strictly convex.
:f(ax+(1-a)y) < af(x)+(1-a)f(y) for a not 0,1.
:I think this is a contradiction to the homogenity of the norm.
:

(For simplicity, Frobenius norm is understood to be considered for real
matrices, x, and it is f(x) = sqrt(trace(x'*x)) where x' is the transpose
of x. Other contexts (complex or infinite dimensional spaces) need
appropriate adjustments.)

In that case, Frobenius norm is a special case of a norm generated by a
positive definite inner product, namely <x,y> = trace(y'*x). The convexity
statements will hold for all such norms.

 If you amend the definition so that it restricts 0 <= a <= 1 , as
customary in all textbooks I've seen, then your observation is correct: it
is false that

   "for all  x, y, and for all a such that  0 < a < 1,
    f(a*x+(1-a)*y) < a*f(x)+(1-a)*f(y)"

as seen when x is a positive multiple of y (contrary to positive
homogeneity of the norm, as you alluded to).

 Strict inequality can be proved for all pairs of linearly independent
x, y. Just square the left side, multiply out and use the strict form of
Cauchy-Schwarz inequality - and in the end, take square roots,
appreciating the assumption 0 <= a <= 1.

:But the SQUARE of the frobenius has the chance of being strictly convex.
:Can someone proof or disproof, please?
:I don't know how. Thank you very much.
:
 Still, strictly speaking, one has to assume that x is different from y,
besides 0 < a < 1.

Then you write f(a*x + (1-a)*y) = f(y + a*(x-y)), square it, multiply out,
and obtain a quadratic function of a. Use second derivative test for
strict convexity.

Good luck, ZVK(Slavek).