Positivity of an operator 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 $f:\mathbb{R}\to \mathcal{M}^{n\times n}(\mathbb{R})$ of class $C^1$, where $\mathcal{M}^{n\times n}$ are real $n\times n$ 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?
 A: 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
$$ M_t (x_1,x_2, \cdots, x_n, \cdots ) = \left(\varphi _1(t) x_1, \frac{\varphi_2(t)}{2} x_2, \cdots, \frac{\varphi_n (t)}{n} x_n, \cdots\right).$$
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 $f'(0) = M$.
