Decoupling of degrees of freedom in Klein-Gordon equation In David Tong's notes in QFT he states that the degrees of freedom decouple in momentum space for the Klein-Gordon eq. He writes that this can be seen by using the Fourier transform (see picture below). I've tried to reproduce this below, but I cannot show the last equality. Does anyone have an idea on what to do? 
\begin{align}
0&=\left(\partial_\mu\partial^\mu+m^2\right)\phi(\vec{x},t)\\
&=(\partial^2_t-\partial_i\partial^i+m^2)\phi(\vec{x},t)\\
&=(\partial^2_t+m^2)\phi(\vec{x},t)-\int_{-\infty}^{\infty}\frac{d^3p}{(2\pi)^3}
\partial_i\partial^ie^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)\\
&=(\partial^2_t+m^2)\phi(\vec{x},t)-\int_{-\infty}^{\infty}\frac{d^3p}{(2\pi)^3}
(ip^i)\partial_ie^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)\\
&=\int_{-\infty}^{\infty}\frac{d^3p}{(2\pi)^3}\left(\partial^2_t+p^2+m^2\right) 
e^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)\\
&\stackrel{??!}{=}\left(\partial^2_t+p^2+m^2\right) 
\phi(\vec{p},t)\\
\end{align}


 A: I also struggled with this point, but below is the solution I ended up with, which uses another Fourier transform.
Start from the line where you got stuck:
$$
     \int_{-\infty}^\infty \frac{d^3\vec{p}}{(2\pi)^3} \left( \partial_t^2 + \vec{p}^2 + m^2 \right) e^{i\vec{p}\cdot\vec{x}} \phi(\vec{p},t) = 0.
$$
This expression says a function of $\vec{x}$ is zero for all values of $\vec{x}$. Since it is a function we can apply an inverse Fourier transform to both sides of the equation, making it a function of a new momentum variable $\vec{q}$:
$$
\begin{aligned}
     \int_{-\infty}^\infty d\vec{x} \, e^{-i\vec{q}\cdot\vec{x}} \int_{-\infty}^\infty \frac{d^3\vec{p}}{(2\pi)^3} \left( \partial_t^2 + \vec{p}^2 + m^2 \right) e^{i\vec{p}\cdot\vec{x}} \phi(\vec{p},t) &= 0 \\
     \Rightarrow \int_{-\infty}^\infty d\vec{p} \, \left( \partial_t^2 + \vec{p}^2 + m^2 \right) \phi(\vec{p},t) \left(\int_{-\infty}^\infty d\vec{x}^3 \frac{e^{i(\vec{p}-\vec{q})\cdot\vec{x}}}{(2\pi)^3} \right) &= 0 
\end{aligned} 
$$
Recognising that the term in parenthesis is a definition of the delta function, we can write:
$$
\begin{aligned}
     &\Rightarrow \int_{-\infty}^\infty d\vec{p} \, \left( \partial_t^2 + \vec{p}^2 + m^2 \right) \phi(\vec{p},t) \delta^3(\vec{p}-\vec{q}) = 0 \\
     &\Rightarrow \left( \partial_t^2 + \vec{q}^2 + m^2 \right)  \phi(\vec{q},t) = 0 \qquad \forall\vec{q} \\
     &\Rightarrow \left(\partial_t^2 + \vec{p}^2 + m^2\right)\phi(\vec{p},t) = 0 \qquad \forall\vec{p}.
\end{aligned}
$$
Which is explicit but rather brute-force. I think the authors were expecting us to recognise the first line is already in the form $\mathcal{F}[f(\vec{x})](\vec{q}) = 0$ and directly conclude that $f(\vec{x}) = 0$.
