So we have that
$$ g(t) = \frac{1}{T}\int_{t-T}^{t}f(\tau) d\tau $$ for $T>0$ and I'm asked to show that $\left| \hat{g}(w) \right|≤\left| \hat{f}(w) \right|$. The hint I get from the question is that I should have convolution in mind.
So I've narrowed it down to this:
We have a convolution between $f(t)$ and some other function, say $h(t)$, that we define to be $\frac{1}{T}$ in some interval and $0$ otherwise. I'm familiar with the Heaviside function and such but I can't make it work in my head. So finally I peaked in the solution for the test and they did it like this:
They defined
$$ h(t) = \begin{cases} \frac{1}{T} & 0<t<T \\ 0 & everywhere \, else \end{cases} $$ and then we can write our integral like this
$$ g(t) = \int_{-\infty}^{\infty}h(t-\tau)f(\tau)d\tau $$ and this I simply don't understand. I have done quite a few of these integrals including both "normal" functions and some easier Heaviside functions, but I can't derive the integral in the question from this last one and I can understand why the limits in $h(t)$ are stated as it is? Is there anyone that can explain this for me?
For the given function $h(t)$ you have $h(t-\tau)=1/T$ for $0<t-\tau <T$, which is equivalent to $\tau<t$ and $\tau > t-T$, and these are exactly your integration limits. And since $|\hat{h}(w)|$ is (the magnitude of) a sinc function with maximum equal to $1$, the given inequality follows.