Determining whether an operator is positive or not Let $T=\int_a^bT_w dw$, where each $T_w$ is a strictly positive, self-adjoint, compact operators acting on a separable Hilbert space $X$. Then, I wonder whether T is also a strictly positive operator. That is, its eigenvalues are all strictly positive. I am naturally guessing that the answer is yes, but I cannot prove it. Above all, I do not know how rigorously one can define the operator $T$ given by an integral of operators, which I think should be the starting point of the proof. I am thinking that the other steps would be just simply straightforward if only it is defined rightly. Also, any reference related to this kind of material would be appreciated. Thanks.
 A: I'd say that the most general integral one usually considers for operators on a Hilbert space is the weak integral, in
the following sense: A function
$$
  w\in [a, b]\mapsto T_w\in B(H)
  $$
is weakly integrable if,  for every vectors $\xi $ and $\eta $ in $H$,  the map
$$
  w\mapsto \langle T_w(\xi ), \eta \rangle
  $$
is Lebesgue integrable on $[a, b]$.  In this case it may be shown that there exists a unique bounded operator $T$ such that
$$
  \int_a^b \langle T_w(\xi ), \eta \rangle \,  dw = \langle T\xi , \eta \rangle ,
  $$
for every $\xi $ and $\eta $ in $H$.
All of this immediately generalizes if the interval $[a,b]$ is replaced by  a general  measure space.
This integral is more general than both Bochner's and Pettis'.  As @Ben Grossmann noticed, a Bochner integral with a
compact integrand gives a compact operator, but this property is not shared by either the weak or the Pettis integral.
If $T_w$ is weakly integrable and each $T_w$ is strictly positive, then so is $\int_a^b T_w\, dw$ by the very same
argument as in @Ben's answer.

EDIT (Answering @Ben Grossmann's comment)
There is a bit of a conflict surrounding the word "weak" in operator theory, because what is universally known as the
weak operator topology (WOT) on $B(H)$ has little to do with its weak topology (W) as a Banach space.
In fact, $B(H)$ is a dual space, namely it is the dual of the space $L^1(H)$ of trace class operators, so $B(H)$ also
has a weak$^*$ topology (W$^*$) arising from the duality
$$
  (T,S)\in L^1(H)\times B(H)\mapsto  \text{tr}(TS)\in \mathbb C.
  $$
The functionals defining the WOT, namely
$$
  T\in B(H)\mapsto \langle T\xi , \eta \rangle \in \mathbb C,
  \tag 1
  $$
for $\xi ,\eta \in H$,
are in fact W$^*$-continuous, so the hierarchy is
$$
  \text{WOT} ⊆  \text{W}^* ⊆  \text{W},
  $$
(all inclusions being proper).
While  the expression "weak operator theory" motivated the terminology   "weak
integral" I used in my answer, a Banach space theorist would perhaps be more inclined to use a term involving the expression "weak$^*$".
Terminology apart, it is conceivable that a function $f$ from a measure space $(\Omega , \mu )$ to $B(H)$ is such that the integral
$$
  \int_\Omega \varphi (f(x))\,d\mu(x)
  $$
is well defined for every $\varphi $ in the dual space $B(H)'$, and yet $f$ is not Pettis integrable.
En passant,  in this case,  one can find a vector $\Lambda $ in the double dual space $B(H)''$ such that
$$
  \int_\Omega \varphi (f(x))\,d\mu(x) = \Lambda (\varphi ), \quad\forall \varphi \in B(H)'.
  $$
The lack of the Pettis integrability of $f$ will be precisely expressed by the fact that $\Lambda $ lies not in the canonical
copy of $B(H)$ within $B(H)''$.
The above  hypothetical function  $f$  will of course satisfy (1),  so it is weakly integrable (WOT) and its weak integral will be the operator
$T$ in $B(H)$ obtained as the image  of $\Lambda $ under the natural projection
$$
  B(H)'' = L^1(H)''' \to   L^1(H)' = B(H).
  $$
Concluding, I am a bit embarrassed not to be able to give an explicit example of a function $f$ with the above
properties but  I am basing my hopes of finding it on the fact that such non-Pettis integrable functions
exists in many non-reflexive spaces.  On the other hand, I might be totally wrong and, in this case, proving myself
wrong will surely lead to a worthwhile result!
A: I will assume that the integral is defined to be the Bochner integral.
An operator $T \in B(H)$ is strictly positive iff for all $x \in H$, we have $\langle Tx, x \rangle > 0$. We note that for any $x \in H$, we have
$$
\langle Tx,x \rangle = \left\langle \left(\int_a^b T_w dw\right)x,x\right\rangle
= \int_a^b \langle T_wx,x \rangle\,dw.
$$
Because each $T_w$ is strictly positive, we have $\langle T_wx,x \rangle > 0$ for all $w$. By the properties of Lebesgue integration, we can conclude that $\int_a^b \langle T_wx,x \rangle\,dw > 0$. So, it indeed holds that $\langle Tx, x \rangle$ is positive for all $x \in H$.
So, $T$ is necessarily a strictly positive operator.
