Change of variables in proof of Bochner's theorem

Let's prove it backwards. We have a function $A(y)=e^{-iyx}\phi(y)$ which we want to integrate over a region. $$ \int_0^T \int_0^T A(t-s)dt ds $$

The integrating region is a square $[0,T]\times [0,T]$, however our function has certain symmetry, it is not any function in two variables $t,s$ but a function in one variable $t-s$. This means that it is constant on level curves of $C(t,s)=t-s$, i.e., on lines of slope ${45}^\circ$.

Let's introduce a change of variables $$ u = t - s\\ v = t + s $$ We have $dtds =\tfrac{1}{2} dudv$. The variables are constrained by $u\in[-T,T]$ and $v\in [|u|,2T-|u|]$, you can see this by drawing the region over which you integrate. Therefore after change of variables we obtain $$ \int_0^T \int_0^T A(t-s)dt ds = \int_{-T}^T \int_{|u|}^{2T-|u|}A(u) \tfrac{1}{2}dvdu =\\ \int_{-T}^T \left(\int_{|u|}^{2T-|u|} \tfrac{dv}{2}\right )A(u)du = \int_{-T}^T (\tfrac{2T - 2|u|}{2})A(u)du = T\int_{-T}^T\left(1-\frac{|u|}{T}\right) A(u)du. $$

Plugging in the correspoiding limits and constant, dividing everything by $T$ and renaming $u$ to $t$ gives the full solution.

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.

Let $s\in\mathbb{R}$ (fixed). Start by $\displaystyle{\frac{1}{2\pi}\int_{-T}^T\bigg{(}1-\frac{|t|}{T}\bigg{)}e^{-itx}\phi(t)dt}$ and change the variable to $t=t'-s$. Then you get the expression $\displaystyle{\frac{1}{2\pi}\int_{-T-s}^{T-s}\bigg{(}1-\frac{|t-s|}{T}\bigg{)}e^{-i(t-s)x}\phi(t-s)dt}$ (note that since $t'$ is a dummy variable i simply wrote $t$ instead, its the same thing). This is true for any $s\in\mathbb{R}$, so as a function of $s$, this is constant. Integrate with respect to $s$ over $[0,T]$; thus this integral is equal to $T\cdot\frac{1}{2\pi}\int_{-T}^T(1-\frac{|t|}{T})e^{-itx}\phi(t)dt$ (we simply integrated a constant function over an interval!), so we have $$ \frac{1}{2\pi}\int_{-T}^T\bigg{(}1-\frac{|t|}{T}\bigg{)}e^{-itx}\phi(t)dt=\frac{1}{2\pi T}\int_0^T\int_{-T-s}^{T-s}\bigg{(}1-\frac{|t-s|}{T}\bigg{)}e^{-i(t-s)x}\phi(t-s)dtds$$

Now I am almost sure that breaking the inside integral at $s$ to lose the absolute value will yield the desired expression. EDIT: apparently this breaking the integral won't work. But draw the image of the region of the plane over which you are integrating. it is a parallelogram with edges $(T,0), (-T,0), (-2T,T), (0,T)$. Look for a linear transformation that will transform this to the square $[0,T]\times [0,T]$ and apply this as a change for variables. This should work.