Uniform Combs of Hat functions

$ \def \half {\frac{1}{2}} \def \hieruit {\quad \Longrightarrow \quad} $ The subject of our current study are uniform combs $P(x)$ of normed hat shaped curves / functions $p(x)$:

Such combs are defined as: $$ P(x) = \sum_{L=-\infty}^{+\infty} p(x-L.\Delta)\,\Delta $$ Where $\Delta$ is the discretization interval length and where being normed means that: $$ \int_{-\infty}^{+\infty} p(x)\,dx = 1 $$ In addition, all hat functions are assumed to be symmetrical around $x = 0$: $$ p(-x) = p(x) $$ Given a sufficient refinement of the discretization $\Delta$ - to be defined later - the comb $P(x)$ can be interpreted as a Riemann sum, approximating the following integral, with $\Delta \rightarrow d\xi$. This explains the factor $\Delta$ in the above definition. $$ \lim_{\Delta \rightarrow 0} P(x) = \int_{-\infty}^{+\infty} p(x-\xi)\,d\xi = 1 $$ The function $P(x)$ can be interpreted as an attempt to "smooth" the uniform density $x_L = L.\Delta$. Or "make fuzzy" the discretization $f(x_L) = 1$ of a constant - and continuous - function $f(x) = 1$. This could be called the Special Theory of Continuity.
It is easily shown that the above function $P(x)$ is periodic. Its period is equal to $\Delta$: $P(x + \Delta) = P(x)$ for arbitrary $x$. Meaning that $P(x)$ can be developed into a Fourier series. The Fourier series of any periodic function is given by: $$ P(x) = \half a_0 + \sum_{k = 1}^{\infty} a_k \, cos(k \omega x) $$ But, in addition, the function is even, meaning that $P(x)=P(-x)$, which results in real-valued Fourier coefficients $a_k$ : $$ a_k = \frac{1}{\Delta/2} \int_{-\Delta/2}^{+\Delta/2} P(x) \cos(k \: 2 \pi / \Delta \, x) \, dx $$ In the sequel, kind of an angular frequency $\omega$ will stand for the quantity $ = 2\pi/\Delta$. Then let the calculations continue: $$ = \frac{1}{\Delta/2} \int_{-\Delta/2}^{+\Delta/2} \sum_{L = -\infty}^{+\infty} p(x - L \Delta)\,\Delta \cos(k\omega x) \, dx $$ $$ = 2 \times \sum_{L = -\infty}^{+\infty} \int_{-\Delta/2}^{+\Delta/2} p(x - L \Delta) \cos(k\omega x) \, dx $$ Substitute $y = x - L \Delta$ and integrate to $y$: $$ a_k/2 = \sum_{L = -\infty}^{+\infty} \int_{- \Delta/2 - L \Delta}^{+ \Delta/2 - L \Delta} p(y) \cos(k \omega [y + L \Delta]) \, dy $$ Where: $$ \cos(k\omega [y + L \Delta]) = \cos(k\omega y + k.L.2\pi) = \cos(k\omega y) $$ Next replace $y$ by $-y$ and switch integration bounds: $$ a_k/2 = \sum_{L = -\infty}^{+\infty} \int_{L \Delta - \Delta/2}^{L \Delta + \Delta/2} p(y) \cos(k \omega y) \, dy $$ The above integrals are precisely the adjacent pieces of another integral which has bounds reaching to infinity. That is, they sum up to an infinite integral: $$ a_k/2 = \int_{- \infty}^{+ \infty} p(y) \cos(k \omega y) \, dy $$ Now the (continuous) Fourier integral of $p(x)$ is defined by: $$ A(y) = \int_{- \infty}^{+ \infty} p(x) \cos(x y) \, dx $$ Wherefrom it is concluded that the (discrete) coefficients of the Fourier series are a sampling of the (continuous) Fourier integral: $$ a_k/2 = A(k \omega) $$ And especially: $$ a_0/2 = A(0) = \int_{- \infty}^{+ \infty} p(x) \, dx \hieruit \half a_0 = 1 $$ Therefore the general expression for the Fourier series of a uniform comb of hat functions is: $$ P(x) = 1 + 2 \times \sum_{k = 1}^{\infty} A(k \omega) \, cos(k \omega x) $$ Where it is reminded that $\omega = 2\pi / \Delta$. And: $$ A(y) = \int_{- \infty}^{+ \infty} p(x) \cos(x y) \, dx $$ It is seen herefrom that $P(x)$, indeed, is an approximation of the constant function $f(x) = 1$, provided that the rest of the Fourier series is just a minor correction on this value. At hand of a few sample hat functions $p(x)$, we will investigate if such is the rule, rather than an exception.