The physical interpretation of the weak energy condition (WEC) I am confused by a statement in Sean Carroll's "Spacetime and geometry". 
On page 174-175, he makes the following statement about the physical interpretation of the WEC in the case of a perfect fluid:

"Because the pressure is isotropic, $T_{\mu\nu}t^\mu t^\nu$ will be non-negative for all timelike vectors $t^\mu$ if both $T_{\mu\nu}U^\mu U^\nu \geq 0$ and $T_{\mu\nu}l^\mu l^\nu \geq 0$ for some nullvector $l^\mu$ ".

$T_{\mu\nu}$ is the energy momentum tensor and $U^\mu$ is the four velocity. For a perfect fluid, the two last inequalities respectively reduce to $\rho \geq 0$ and $\rho + p \geq 0$.
I do understand that both inequalities are necessary conditions for the WEC ( $T_{\mu\nu}t^\mu t^\nu \geq 0$) to hold, but they are not obviously sufficient conditions. 
They are necessary conditions as $U^\mu$ is a timelike vector and, by continuity, the inequality involving $l^\mu$ should hold as well. However, they are treated as sufficient conditions, i.e. the WEC is used as a synonym for "$\rho \geq 0$ and $\rho + p \geq 0$".
In a metric that does not differ too much from a Minkowskian metric (weak field limit), we can decompose any timelike vector $t^\mu$ as a sum of a well chosen nullvector $l^\mu$ and a multiple of $U^\mu$ or $$t^\mu = l^\mu+ \lambda U^\mu$$ with $\lambda$ a real number. This directly leads to  $$T_{\mu\nu}t^\mu t^\nu = \rho (\lambda -U_\mu l^\mu )^2 +p(U_\mu l^\mu)^2$$. This last equation implies that in the weak field limit, the WEC is equivalent to $\rho \geq 0$ and $p \geq 0$. Can somebody help me understanding where is the mistake I am making and why the conditions are sufficient conditions?
 A: You have already noted that they are necessary conditions. That they are sufficient follows precisely because the pressure is isotropic. To see this consider a local Lorentz (orthonormal) frame (a.k.a tetrad). Any null vector can be written as
$$
\ell^i = N\left(\delta^i_0 + \delta^i_1\right),
$$
since we have specified no spacelike directions (really the 1 is just an arbitrarily chosen spacelike index), while
$$
U^i = \delta^i_0.
$$
The stress-energy tensor is $T_{ij} = \mathrm{diag}\left(\rho,p,p,p\right)$. Then 
$$T_{ij}U^iU^j = T_{00} = \rho \geq 0$$
and
$$T_{ij}\ell^i\ell^j = N^2\left(T_{00} + T_{11}\right) = N^2\left(\rho + p\right) \geq 0$$
is equivalent to that $\rho + p \geq 0$. Finally, an arbitrary timelike vector can be decomposed, as you say
$$
t^i = \ell^i + \lambda U^i, 
$$
with $\lambda^2 + 2N\lambda > 0$ (equal to 1 if normalized), since otherwise $t^i$ would be null or spacelike. Thus
\begin{align}
T_{ij}t^it^j &= N^2(\rho + p) + (\lambda^2 + 2N\lambda)\rho \geq 0
\end{align}
by the above results. Meaning that all observers observe the energy density to be non-negative.
A: For future readers of this thread, here is a description of how to show that the decomposition $t^\mu = l^\mu + \lambda U^\mu$, which is used in the accepted answer, always exists (locally).
Let us denote the norm of the timelike vector $t^\mu$ by
$$T = t^\mu t_\mu < 0.$$
We wish to find a null vector $l^\mu$ and a number $\lambda$ such that $t^\mu = l^\mu + \lambda U^\mu$. We will then have
\begin{split}
l^\mu l_\mu &= (t^\mu - \lambda U^\mu)(t_\mu - \lambda U_\mu)\\
&= t^\mu t_\mu - 2 \lambda U^\mu t_\mu + \lambda^2 U^\mu U_\mu.
\end{split}
Suppose $t^0 = S$ in the rest frame of the perfect fluid source of the energy-momentum tensor, i.e. the frame in which $U^\mu = (1,0,0,0)^\mu$. Then
$$l^\mu l_\mu = T + 2 \lambda S - \lambda^2.$$
For $l^\mu$ to be null we need $l^\mu l_\mu = 0$, and hence
$$\lambda^2 - 2 \lambda S - T = 0.$$
This equation can be solved for $\lambda$ provided that $S^2 + T \ge 0$. But, using locally inertial coordinates to evaluate $T$, we find that
$$T = t^\mu t_\mu = -(t^0)^2 + \sum_{i=1}^3 (t^i)^2 \ge - S^2,$$
so it is always possible to solve for $\lambda$.
