# How to prove that the peak of the autocorrelation function is at zero lag?

Show that for a signal $f(\tau)$ with finite energy and energy autocorrelation function $\phi^e_{ff} (\tau),$$|\phi_{ff}^e (\tau)| \leq \phi_{ff}^e (0), \ \ \forall \tau.$$ According to my textbook the total energy is given by$\phi_{ff}^e (0),$but it does not provide any proof. The two energy auto-correlation functions here would become: $$\phi_{ff}^e (\tau) = \int^\infty_{- \infty} f^*(t) f(t+ \tau) \ dt,$$ $$\phi_{ff}^e (0) =\int^\infty_{- \infty} f^*(t) f(t) \ dt.$$ I was told that we must use the Cauchy-Schwarz inequality for this proof. But I am not exactly where to apply it. Any help would be appreciated. ## 1 Answer The Cauchy Schwarz inequality states that: $$\left|\int_{-\infty}^{\infty}g_1(t)g_2(t) dt\right|^2 \leq \int_{-\infty}^{\infty}|g_1(t)|^2 dt \int_{-\infty}^{\infty}|g_2(t)|^2 dt$$ I'm going to assume that$f(t)$is real, just to make the math a little easier. From the above we can write: $$\left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq \int_{-\infty}^{\infty}|f(t)|^2 dt \int_{-\infty}^{\infty}|f(t-\tau)|^2 dt$$ For the second integral on the right, just use a variable substitution - let$u=t-\tau$and then rewrite it with the dummy variable$t$again - we have $$\left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq \int_{-\infty}^{\infty}|f(t)|^2 dt \int_{-\infty}^{\infty}|f(t)|^2 dt$$ but the integral on the right is just the autocorrelation function i.e.$ \int_{-\infty}^{\infty}|f(t)|^2 dt= R_{f}(0)$. So we can write $$\left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq R^2_{f}(0)$$ Note that from the definition of$R_f(0)$it must be a positive quantity, so taking the square root of both sides and paying attention to the positive and negative quantities on the left hand side, we can write $$\int_{-\infty}^{\infty}f(t)f(t-\tau) dt \leq R_{f}(0)$$ Note - you don't have to use the Cauchy Schwarz inequality - another proof: $$[f(t)-f(t-\tau)]^2 = f^2(t)+f^2(t-\tau) -2f(t)f(t-\tau)$$ Integrating both sides from$-\infty$to$\infty$and using the same trick (variable substitution for the integral over$f^2(t-\tau)\$) we can write $$\int_{-\infty}^{\infty} [f(t)-f(t-\tau)]^2dt= 2\int_{-\infty}^{\infty}f^2(t)dt-2\int_{-\infty}^{\infty}f(t)f(t-\tau)dt$$

Rearranging we can write: $$\int_{-\infty}^{\infty}f(t)f(t-\tau)dt = \int_{-\infty}^{\infty}f^2(t)dt -\frac{1}{2}\int_{-\infty}^{\infty} [f(t)-f(t-\tau)]^2dt$$ Now - the second integral on the right has to be a positive quantity and it is subtracted so we have $$\int_{-\infty}^{\infty}f(t)f(t-\tau)dt \leq \int_{-\infty}^{\infty}f^2(t)dt$$