Let $f \in C^0_c(R)$ (continuous function with compact support) and $w \in L^1(R)$ (Lebesgue integrable function) such that $w(x) \geq 0$ for all $x \in R$. The running autocorrelation function (it is given by Licklider in this paper p. 133) is the function:
$$\begin{matrix} \phi : & R^2 & \to & R\\ & (t,\tau) & \mapsto & \displaystyle\phi(t,\tau) := \int^\infty_{0} f(t - T)f(t - \tau - T)w(T) \, dT \end{matrix} $$
I want to prove that $\phi$ is continuous (therefore it spawns a surface in $R^3$ as Licklider says). This is the sketch of my attempt:
Let $((t_k,\tau_k))$ be a sequence in $R^2$ such that $(t_k,\tau_k)\to (a,b)$ for some $(a,b)\in R^2$, I have to show that the sequence $\phi(t_k,\tau_k) \to \phi(a,b)$ in $R$, that is, I have to show that
\begin{align} & \phantom{\; \iff \;}\lim_{k\to \infty} \int^\infty_{0}f(t_k - T)f(t_k - \tau_k - T)w(T) \, dT = \int^\infty_{0}f(a - T)f(a - b - T)w(T) \, dT \end{align}
It is very tempting to move the limit inside the integral and we would have that:
$$\lim_{k\to\infty} f(t_k - T)f(t_k-\tau_k - T)w(T) = f(a - T)f(a - b - T)w(T)$$
since $f$ is continuous and $t_k \to a$ and $\tau_k \to b$, and we would be done. To be sure I can indeed do that I used the dominated convergence theorem, and defined the sequence $(h_k)$ of functions such that $h_k(T) = f(t_k - T)f(t_k-\tau_k - T)w(T)$, we would have pointwise convergence as: $\lim_{k \to \infty} h_k(T) = h(T)$ where $h(T) := f(a - T)f(a - b - T)w(T)$ for all $T$, however it feels weird to define a sequence such as $h_k(T)$, I don't know if this can be done or if there is a problem with that. Then for the function that dominates every element of the sequence we could take $M$ the maximum of $|f|$ (compact support means it has a maximum) and define $g(T) = M^2w(T)$, we have that $g$ is integrable and it fulfills that $|h_k(T)| \leq g(T)$ for all $h_k$, then we are good to use the theorem and prove the continuity.
¿Is there something wrong with this attempt or is there something I am missing?