Integration of autocorrelation function when $\int_0^1 f(t) dt = 0$

Question

When $\int_0^1 f(t) dt = 0$ and $f(t)=0$ for $t\in \mathbb{R} \setminus [0,1]$, my conjecture is that $$ \int_{\mathbb{R}} R_{ff}(\tau)\,d\tau = 0, $$ where $$ R_{ff}(\tau) = \int_{\mathbb{R}} f(t+\tau) f(t)\,dt. $$ How can I prove this?

Answer 1

Since $f(t) = 0$ for $t>1$ or $t<0$, it is enough to show \begin{align*} \int_{-C}^C R_{ff}(\tau)\,d\tau, \end{align*} for some $C>1$. By the integration by parts, \begin{align*} \def\bbR{{\mathbb{R}}} \require{cancel} \int_{-C}^C R_{ff}(\tau)\,d\tau &= \cancelto{0}{\tau R_{ff}(\tau)\rvert_{-C}^C} - \int_{-C}^C \tau R'_{ff}(\tau)\,d\tau \\ &=- \int_{-C}^C \tau \int_{\bbR} f'(t+\tau) f(t)\,dt \,d\tau \\ &=- \int_{\bbR} \int_{-C}^C \tau f'(t+\tau)\,d\tau f(t)\,dt \\ &=- \int_{\bbR} \left( \cancelto{0}{\tau f(t+\tau) \rvert_{-C}^C} - \cancelto{0}{\int_{-C}^C f(t+\tau)\,d\tau} \right) f(t)\,dt \\ &=0. \end{align*}