Does the Lorentz invariance of equation of motion guarantee the Lorentz invariance of the solutions? If I have a Lorentz invariant equation of motion, like Klein-Gordon equation, is the solution automatically guaranteed to be Lorentz invariant?
I ask this question because of the discussion from Mark Srednicki's Quantum Field Theory section 3 from equations (3.11) to (3.14). If I have a K-G equation, 
$$ \tag 1 \partial^\mu\partial_\mu\phi -m^2\phi=0,$$
we have a solution of the form
$$ \tag 2\exp (i \mathbf{k} \cdot \mathbf{x} \pm i\omega t),$$
which I do not think is Lorentz invariant for solution with
$i \mathbf{k} \cdot \mathbf{x} + i\omega t$
as an argument, unless we allow $k^\mu = (-\omega, \mathbf{k})$.
However, he starts constructing a Lorentz invariant solution, and comes up with 
$$ \tag 3 \phi(\mathbf{x},t) = \int d\tilde{k}[ a(\mathbf{k})e^{ikx} + a^*(\mathbf{k})e^{-ikx}],$$
where
$kx = \mathbf{k}\cdot\mathbf{x} - \omega t$.
$d\tilde{k}$ is a Lorentz invariant measure and argument of each exponents are Lorentz invariant as well.
However, he says in the beginning that $a(\mathbf{k})$ is an arbitrary function of the wave vector $\mathbf{k}$, which does not sound Lorentz invariant to me. So I am not sure how $\phi(\mathbf{x},t)$ is Lorentz invariant.
 A: In the spirit of the original post, let $k,x$ be 4-vectors and $\mathbf{k}$, $\mathbf{x}$ the spatial components. Then a quantity of the form
$$\phi(x) \propto \int dk[ a(k)e^{ikx} + a^*(k)e^{-ikx}]$$
is manifestly Lorentz invariant because it does not explicit contain any free Lorentz indices. What Srednicki does is that he performs the $k^0$ integration, resulting in
$$\phi(\mathbf{x},t) = \int \frac{d\mathbf{k}}{f(\mathbf{k})}[ a(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}} + a^*(\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}}],$$
which only includes spatial components. This expression is Lorentz invariant because it is just a different form of the previous one, but it does not manifestly look Lorentz invariant which I assume what causes the confusion. For an explicit form of the function $f$ which will of course be related to the energy as it is the integral over $k^0$, see for example Peskin and Schroeder eqn (2.47).

EDIT: Some more justification:
The Klein-Gordon eqn is
$$\partial^\mu\partial_\mu\phi -m^2\phi=0.$$
To solve it, we Fourier transform to momentum space and we get:
$$(p^\mu p_\mu -m^2)\tilde\phi=0.$$
The general solution of this eqn is
$$\tilde\phi(p)=a(p)\delta(p^\mu p_\mu -m^2),$$
which means that the general solution for the Klein-Gordon is:
$$\phi(x)=\frac{1}{(2\pi)^4}\int d^4pe^{ipx}\tilde\phi(p)=\frac{1}{(2\pi)^4}\int d^4pe^{ipx}a(p)\delta(p^\mu p_\mu -m^2)$$
which is manifestly Lorentz invariant. You can then perform the $p^0$ integration as claimed above. I have ignored the complex conjugate term everywhere but should be trivial to restore...
