Prove that $\lim_{n\to\infty}\int_0^{2\pi}|f(x-t)\psi_n(t)|\, \text{d}t=f(x)$. Let $ \psi_n: \Bbb R \to \Bbb R $ be a sequence of continuous and periodic functions in $ \Bbb R $  that satisfies $$\int_0^{2\pi}\psi_n(t)dt=1\quad \text{and}\quad \int_0^{2\pi}|\psi_n(t)|dt\leq P$$
for all $n\in\Bbb Z^+$, and $$\lim_{n\to\infty}\int_{\delta}^{2\pi-\delta}|\psi_n(t)|dt=0$$ for each $0<\delta<2\pi$. Suppose that $f:\Bbb R\to \Bbb R$ is continuous and periodic. Prove that
$$\lim_{n\to\infty}\int_0^{2\pi}|f(x-t)\psi_n(t)|\, \text{d}t=f(x).$$
Any help to do this exercise please. I do not know how to do it. Thank you.
 A: As someone has already pointed out in the comments, there is probably a mistake in the thesis, as without assuming $f(x)$ positive you might have a contradiction.
I will prove that
$$
\lim_{n\to\infty} \int_0^{2\pi} f(x-t)\psi_n(t)dt = 0.
$$
I will also assume that all the functions are periodic with period $2\pi$. It wasn't specified, but I figured it had to be so.
For $n$ big enough, to be chosen later, let us split the integral into three pieces
$$
\int_0^{2\pi} f(x-t)\psi_n(t)dt = \int_0^{\delta} (...)dt + \int_{\delta}^{2\pi-\delta}(...)dt + \int_{2\pi - \delta}^{2\pi}(...)dt = I_1 + I_2 + I_3
$$
We focus on $I_2$. Since we know that
$$
\lim_{n\to\infty} \int_{\delta}^{2\pi-\delta}|\psi_n(t)|dt = 0, \ \forall \delta \in (0,2\pi),
$$
and since, being a continuous periodic function, $f$ is bounded on $[0,2\pi]$, we have that for every $\varepsilon >0$ there exists $\bar n$ big enough that
$$
|I_2| \leq \|f\|_{\infty} \varepsilon \ \ \forall n > \bar n.
$$
For what concerns $I_1 + I_3$, emplyoing the fact that $f$ is continuous, we have that for every $\varepsilon >0$ there exists $\bar \delta$ such that
$$
f(x)-\varepsilon < f(x-t) < f(x) + \varepsilon \ \ \forall t < \delta.
$$
Furthermore, being $f$ periodic
$$
f(x)-\varepsilon < f(2\pi+x-t) < f(x) + \varepsilon \ \ \forall t < \delta.
$$
Finally, we have
$$
(f(x)-\varepsilon)\left(\int_0^{\delta}\psi_n(t)dt + \int_{2\pi-\delta}^{2\pi} \psi_n(t)dt\right) < I_1 + I_3 < 
(f(x)+\varepsilon)\left(\int_0^{\delta}\psi_n(t)dt + \int_{2\pi-\delta}^{2\pi} \psi_n(t)dt\right)
$$
for every $\delta < \bar \delta$ small enough. Since by hypothesis
$$
1>\int_0^{\delta}\psi_n(t)dt + \int_{2\pi-\delta}^{2\pi} \psi_n(t)dt > 1-\varepsilon,
$$
we have proved that for every $n > \bar n$
$$
(1-\varepsilon)(f(x)-\varepsilon) - C\varepsilon < I_1 + I_2 + I_3 < (f(x)+\varepsilon) + C\varepsilon,
$$
which is the thesis.
