Positivity of an operator

Question

Consider a function $f:mathbb{R}to mathbb{R}$ of class $C^1$. If $f(0)=0$ and $f'(0)>0$ it's clear that there exist some $t_0>0$ such that $f(t_0)>0$.
Now if $f:mathbb{R}to mathcal{M}^{ntimes n}(mathbb{R})$ of class $C^1$, where $mathcal{M}^{ntimes n}$ are real $ntimes n$ matrices, if $f(0)=0$ and if $f'(0)$ is a strictly positive defined matrix, again there will be a $t_0$ such that $f(t_0)$ is a strictly positive defined matrix.
The question is, is it true even for operators? In particular, let $f:mathbb{R}to mathcal{O}$ of class $C^1$, where $mathcal{O}$ is the set of compact self-adjoint operators on some separable Hilbert space $mathcal{H}$. Let $f(0)=0$ and suppose that $f'(0)$ is a compact positive self-adjoint operator, is it true that there must be a $t_0$ such that $f(t_0)$ is positive?

Arctic Char · Accepted Answer

No. Counterexample: Let $H = ell^2$ and $M : H to H$ be given by
$$ M(x_1, x_2, cdots, x_n , cdots) = left( x_1, frac{x_2}{2}, cdots, frac{x_n}{n}, cdots right).$$
Then $M$ is compact (limits of finite rank operators), self-adjoint and positive. Next let $varphi: mathbb R to mathbb R$ be a smooth odd function so that

$varphi(t) = t$ on $[-1,1]$,
$|varphi (t)|le 1.1$
$varphi$ is decreasing on $[1.1, 2]$ and
$ varphi(t) = 0$ on $[2, infty)$.

For each $n$, define $varphi_n (t) = frac{1}{2^n }varphi (2^n t)$. Define $ M_t:=f(t)$ by
$$ M_t (x_1,x_2, cdots, x_n, cdots ) = left(varphi _1(t) x_1, frac{varphi_2(t)}{2} x_2, cdots, frac{varphi_n (t)}{n} x_n, cdotsright).$$
Then $M_0 = 0$ and each $M_t$ is self-adjoint, finite rank (thus non-positive). Also, $f$ is $C^1$. Indeed one can check that
$$f'(t) (x_1,x_2, cdots, x_n, cdots ) = left( varphi_1'(t) x_1, frac{varphi_2'(t)}{2} x_2, cdots, frac{varphi_n'(t)}{n} x_n, cdots right).$$
Since $varphi_n'(0)=1$ for all $n$, we have $f'(0) = M$.

Positivity of an operator

One Answer

Add your own answers!

Ask a Question