So here is my problem,
I have to solve the following exercise,
Let $\phi\in L^1[0,1)$ and $\psi\in L^{\infty}[0,1)$, both of period 1 and $\int_0^1\psi(t)dt=0$. Show that $$\lim\limits_{n\rightarrow \infty}\int_0^1\phi(t)\psi(nt)dt=0$$
I composed a proof and I somehow do not see why $\int_0^1\psi(t)dt=0$ is necessary. Since usually not using an assumption in a proof means that there is a mistake in it I think that my solution is wrong. But I am not able to find the mistake. Thus I hoped someone could look over my solution.
So here is my proof:
Remark: In the previous exercise I showed that for a $\phi\in L^1[0,1]$ of period 1 the following holds $$\int_0^1 e^{i2\pi kt}g(nt)dt=0$$ for $n>|k|$.
Consequently for a trigonometric polynomial $P(x)$ of degree strictly less than $n$ we have $$\int_0^1 P(x)g(nt)dt=0$$
Back to the actual proof, note first that since for some $\psi\in L^{\infty}[0,1)$ we have, $$\int_0^1|\psi(x)|dx\leq\int_0^1\sup_{x\in[0,1]}|\psi(x)|$$ it follows that $ L^{\infty}[0,1)\subset L^1[0,1)$. Hence, $$\int_0^1P(t)\psi(nt)dt=0$$ for all trigonometric polynomials with degree strictly less then $n$.
Now, let $\phi$ be a 1 periodic and integrable function. By density of trigonometric polynomials in in 1 periodic $L^1$ functions it follows that there exists a trigonometric polynomial $P$ s.t $||\phi-P||_{L^1[0,1)}<\epsilon$ for an arbitrary small $\epsilon>0$. Moreover we have that $\mathrm{deg}(P(x))=k$ for some positive integer $k$. In particular this means that choosing $n\in\mathbb N$ with $n>k$ yields, \begin{align*} |\int_0^1\phi(t)\psi(nt)dt|&=|\int_0^1\phi(t)\psi(nt)dt-\int_0^1P(t)\psi(nt)dt|\\ &\leq \int_0^1|\psi(nt)||\phi(t))-P(t)|dt\\ &\leq \sup_{t\in[0,1]}|\psi(t)||\phi(t))-P(t)|dt\\ &<||\psi||_{\infty}\epsilon\\ \end{align*} Hence for $n$ large enough $$\int_0^1\phi(t)\psi(nt)dt=0$$ and conequently $$\lim\limits_{n\rightarrow \infty}\int_0^1\phi(t)\psi(nt)dt=0$$ $\square$
No you haven't. The assumption in that exercise excluded $k=0$. When $k=0$, you have $$\int_0^1 e^{i2\pi kt}g(nt)dt=\int_0^1 g(x)\,dx$$ This is why the extra assumption is made here: the integral of $g$ (which is $\psi$ in your proof) is zero.
Otherwise, your proof works.