TITLE: Change of variables in proof of Bochner's theorem
QUESTION [6 upvotes]: I have $\phi: \mathbb{R} \to \mathbb{R}$ continuous, bounded, and integrable. I'm going through a book which makes the following calculation:
$$\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}e^{-itx}\phi(t)dt = \lim\limits_{T\to \infty}
\frac{1}{2\pi}\int\limits_{-T}^{T}\left(1 - \frac{|t|}T \right)e^{-itx}\phi(t)dt = 
\lim\limits_{T\to \infty}
\frac{1}{2\pi T}\int\limits_{0}^{T} \int\limits_{0}^{T} e^{-i(t-s)x}\phi(t-s)dtds.
$$ 
The first equality holds by the Dominated Convergence Theorem and the second is supposed to hold by a change of variables. 
Can someone please show me how this change of variables goes?
Edit: If it helps, this is in the book of Varadhan in the proof of Bochner's theorem. This computation is being made to use the positive-definiteness of $\phi$.

REPLY [3 votes]: Let's see this integration as a set of convolutions. In fact, if we set
\begin{align}
q(t) &= u(t)-u(t-T),\\
q'(t)&=q(-t)
\end{align}
where $u(\cdot)$ is the heaviside step function, and
\begin{equation}
p(t)= {\rm e}^{-itx} \phi(t).
\end{equation}
Hence, one can easily check that
\begin{equation}
\Big(1-\frac{|t|}{T}\Big) \big(u(t+T)-u(t-T)\big) = \frac{1}{T}(q'*q)(t),
\end{equation}
in which $*$ is the convolution sign. As a result, one can see that
\begin{equation}
\int_{-T}^T \Big(1-\frac{|t|}{T}\Big) {\rm e}^{-itx} \phi(t){\rm d} t = \frac{1}{T}(q'*q*p)(0).
\end{equation}
Next, using the definiton of $q(t)$, you can rewrite the above formula to see that
\begin{equation}
(q*p)(t) = \int_{s=-\infty}^{\infty} \big(u(s)-u(s-T)\big) p(t-s){\rm d} s = \int_{s=0}^{T} p(t-s){\rm d} s.
\end{equation} 
Finally, by associativity of convolution, we have that
\begin{equation}
\frac{1}{T}(q'*q*p)(0) = \int_{t=0}^T \int_{s=0}^{T} p(t-s){\rm d} s{\rm d} t,
\end{equation}
which completes the proof.