Limiting Property of an Integral? I have observed an interesting property which could be wrong as it seems completely absurd but the motivation behind it makes me inclined to believe it is true (or at least in some cases). I do not have any pure mathematical background so I do not know what I am talking about but I wish this kind community would get the gist of what I am trying to ask and would kindly direct me to an answer. So here goes nothing.
The Observation:
Suppose $g(t)$ is a periodic, symmetric function if it satisfies the following conditions:

*

*There exists a nonzero constant T such that $g(t+T) = g(t)$. (periodic)

*For all $ \delta \in[0,\frac{T}{2}]$, $g(\frac{T}{2}-\delta) = g(\frac{T}{2}+\delta)$. (symmetric)
Then for any integrable function $f(t)$, $\lim_{w \to \infty} \int f(t)g(wt)dt = \alpha \int f(t)dt$ where $\alpha$ is some constant.
My question is: Is this in any way true?
The Motivation:
One function that satisfies the function $g(t)$ is $sin^2(t)$ to which I find with several computational evidences that $\lim_{w \to \infty} \int f(t)\sin^2(wt)dt = \frac{1}{2} \int f(t)dt$.
Thank you for entertaining this question. If you find this question to be nonsensical, then I apologize in advance.
 A: Yes, this ought to be true with appropriate hypotheses. Here is a sloppy argument ignoring convergence issues etc.; maybe a real analyst can tell us what hypotheses and bounds are needed. The integral $\int f(t) g(wt) \, dt$ is the inner product $\langle f(t), g(wt) \rangle$ on $L^2(\mathbb{R})$. Pass to Fourier transforms: since $g$ is periodic with period $T$, its Fourier transform is a sum of Dirac deltas
$$\hat{g}(\xi) = \sum_{k \in \mathbb{Z}} \hat{g}_k \delta \left( \xi - \frac{2 \pi k}{T} \right)$$
where $\hat{g}_k$ is the Fourier series of $g$, and hence the Fourier transform of $g(wt)$ is
$$\frac{1}{w} \hat{g} \left( \frac{\xi}{w} \right) = \sum_{k \in \mathbb{Z}} \hat{g}_k \delta \left( \xi - \frac{2 \pi k w}{T} \right).$$
Now we can apply the Plancherel theorem, which gives
$$\begin{align*} \langle f(t), g(wt) \rangle &= \sum_{k \in \mathbb{Z}} \hat{f} \left( \frac{2 \pi k w}{T} \right) \hat{g}_k \end{align*}.$$
From here we assume that $f$ is well-behaved enough that its Fourier transform decays at infinity. As $w \to \infty$ it follows that all the terms above except the $k = 0$ term go to zero, and so we expect
$$\lim_{w \to \infty} \langle f(t), g(wt) \rangle \stackrel{?}{=} \hat{f}(0) \hat{g}_0 = \boxed{ \left( \int f(t) \, dt \right) \left( \frac{1}{T} \int_0^T g(t) \, dt \right) }.$$
This can be understood intuitively as follows. As $w \to \infty$, the function $g(wt)$ wiggles so rapidly that from its perspective the function $f(t)$ is approximately locally constant (assuming it is e.g. continuous). If we integrate over a small interval $[t_0, t_0 + \varepsilon]$, $f(t)$ is approximately a constant $f(t_0)$ while $g(wt)$ runs over all values in a period of $g$ many times. The conclusion is that heuristically this integral over a small interval is approximately $\varepsilon f(t_0)$ times the average value $\frac{1}{T} \int_0^T g(t) \, dt$ of $g(t)$ over a period. Summing this over all small intervals we recover the result.
This can pretty easily be turned into a rigorous proof for $f(t)$ a step function (I think $f(t)$ a Lipschitz function with a little more difficulty, maybe $f(t)$ continuous with a little more difficulty again), and from there we ought to be able to pass to more general $f$ via some kind of dominated convergence argument. Note that this is also consistent with your example $g(t) = \sin^2 t$ because $\frac{1}{2}$ is the average value of this function over its period; since $\sin^2 t$ has a finite Fourier series it should be particularly easy to check this case.
A: Lets say we have:
$$\int f(t)g(\omega t)dt=\int_0^t f(x)g(\omega x)dx$$
we can break down $t$ into various different regions as so:
$$t=kT+\varepsilon$$
where:
$$\varepsilon\in[0,T)\quad k\in\mathbb{Z}^+$$
now we can instead represent our integral as:
$$\int_0^t f(x)g(\omega x)dx=\sum_{i=0}^{k-1}\int_{iT}^{(i+1)T}f(x)g(\omega x)dx+\int_{kT}^{kT+\varepsilon}f(x)g(\omega x)dx$$
now making the substitution $y=\omega x\Rightarrow dx=dy/\omega$ we get:
$$\sum_{i=0}^{k-1}\frac1\omega\int_{iT\omega}^{(i+1)T\omega}f(y/\omega)g(y)dy+\frac1\omega\int_{kT\omega}^{(kT+\varepsilon)\omega}f(y/\omega)g(y)dy$$
Now I feel that we can possibly make a connection to the Riemann sum although I am stumped as of now, just a thought...
