Functions $f$ such that $u_n = \int_0^1 f(t) \cos(nt) \, \mathrm d t$ converges to $0$ I am trying to characterise all functions $f:[0,1]\to \Bbb R$ such that the sequence $(u_n)$ converges to $0$ where $u_n = \int_0^1 f(t) \cos(nt) \, \mathrm d t$. I know that a sufficient condition is that $f$ is Lipschitz but I have to find necessary and sufficient conditions. I'm not sure if $f$ needs to be bounded, but I suspect it does since the zeros of $\cos(nt)$ change with $n$. Any ideas would be appreciated.
 A: What you are to show is a special case of the Riemann-Lebesgue lemma. A proof can be found in most textbooks on real analysis, but below is a rough summary of the main ideas that goes into proving this. There are plenty of details (and $\epsilon$'s) left to be filled in to make this into a rigorous proof, but that should not be too hard. We only need to assume that $f$ is Riemann integrable.
First prove it for step-functions. Let $\Pi_N = \{x_0=0 <x_1<\ldots < x_N = 1\}$ be a partition of $[0,1]$ and define $s(x) = \sum_{i=1}^N m_i \chi_{[x_{i-1},x_i)}(x)$ where $m_i$ is some constants and $\chi_A(x)$ is the characteristic function of $A$, i.e. $\chi_A(x) = 1$ if $x\in A$ and $0$ otherwise. For such a function we have 
$$\int_0^1 s(x)\cos(nx){\rm d}x = \frac{1}{n}\sum_{i=1}^Nm_i(\sin(nx_i) - \sin(nx_{i-1})$$
and we see that taking $n$ large enough this can be made arbitrarily small. 
Next use that $f$ is Riemann integrable to find a partition $\Pi_N$, which you can use to define a step-function $s(x) = \sum_{i=1}^N m_i \chi_{[x_{i-1},x_i)}(x)$, such that $\int_0^1 f(x) - s(x){\rm d}x$ is as small as you want.
Finally try to estimate $\int_0^1 f(x)\cos(nx)$ by writing $f(x) = (f(x) - s(x)) + s(x)$ and using the triangle inequality. The integral of $s(x)\cos(nx)$ can as shown above be made small by taking $n$ large enough. The integral of $(f(x)-s(x))\cos(nx)$ is less than the integral of $|f(x)-s(x)|\cdot 1 = f(x) - s(x)$ if we also make sure the $m_i$'s we pick are such that $f(x)-s(x)\geq 0$ (try to use the minimum value of $f$ on each interval to define $m_i$). This integral we already know is small by the choice of partition.
