How can this integral be rewritten with convolutions?

Question

I've got $f:\mathbb{R}\rightarrow\mathbb{R}$ bounded and I'm trying to write `$\mathtt{f}$,' a discrete version of $f$, where each element in the domain takes on the average of the corresponding interval in $f$.

Right now I'm writing $\mathtt{f}:\mathbb{Z}\rightarrow \mathbb{R}$ in the following way:

$$\mathtt{f}[n]=\int_{n-\frac{1}{2}}^{n+\frac{1}{2}} f(x)\ dx$$

However, I think this can be written in terms of convolutions and comb functions:

$$\mathtt{f}[n]=\text{rect}(x)\ast \ (\ \text{comb}(x)\cdot(\text{rect}(x)\star f(x))\ ) \text{ evaluated at }n$$

Is this the same as the first definition?

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$\text{rect}$ is the function which is zero everywhere except in the interval $[-\frac{1}{2},\frac{1}{2}]$ where it is 1.

$\text{comb}$ is an impulse train $\text{(i.e., } \Sigma_{k\in\mathbb{Z}}\delta(x-k) \text{ ).}$

$\star$ denotes convolution.

Answer 1

Actually, from your definition $$ \mathtt{f}[n]=\int_{n-\frac12}^{n+\frac12}f(x)\,\mathrm{d}x $$ and $$ \mathrm{rect}(x)=\left\{\begin{array}{l} 1&\text{if }x\in\left[-\tfrac12,\tfrac12\right]\\ 0&\text{otherwise} \end{array}\right. $$ we have that for $n\in\mathbb{Z}$, $$ \begin{align} \mathtt{f}[n] &=\mathrm{rect}\ast f(n)\\[9pt] &=\int_{-\infty}^\infty\mathrm{rect}(n-x)f(x)\,\mathrm{d}x\\ &=\int_{n-\frac12}^{n+\frac12}f(x)\,\mathrm{d}x\\ \end{align} $$

Answer 2

As far as I understand your notations, $$(comb \ast rect)(x) = 1,$$ $$(f\ast rect(\cdot-n))(x) =\int_{n-1/2}^{n+1/2}f(x-y)dy, $$ so the final convolution equals as a function of $z$ $$\int_{\Bbb R}dx \int_{n-1/2}^{n+1/2}f(x-y)dy = \int_{\Bbb R}f(x)dx.$$