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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.

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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 $$ \int_{-\infty}^{\infty}f(t)f(t-\tau)dt \leq \int_{-\infty}^{\infty}f^2(t)dt $$

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