Consider $f\in L^2(\mathbb{T})$ where $\mathbb{T}=[0,2\pi]$ and further denote by $\mathcal{F}$ the Fourier transform. Define \begin{align*} g(t):=\int^t_0f(s)ds. \end{align*} Can we find a function $u$ such that \begin{align*} g=\mathcal{F}^{-1}(u\cdot\mathcal{F}(f)) \end{align*} holds? I would argue that the indicator function $v(t):=\mathbf{1}_{[0,t]}(t-s)$ does the job, thus we found $u=\mathcal{F}(v)$.
As I am not familiar with Fourier Analysis at all, I would like to have this verified but my main concern would be if this also holds in the discrete setting. So can we find Fourier multiplier $P$ such that \begin{align*} g_N = \mathcal{F}^{-1}_N(P\mathcal{F}_N(f_N)) \end{align*} where $\mathcal{F}_N$ is the discrete Fourier Transform and $g_N\in L^2_N(\mathbb{T})$ denotes the space of trigonometric polynomials where we can make use of the Fourier $L^2$ projection on $g$ and $f$.
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$g(t) = \mathcal{F}_{t}^{-1}(G(y))$, where $G(y) = \mathcal{F}_{y}(g(t)) = \int\limits_{-\infty}^{\infty} g(t) e^{-i 2 \pi y t} dt$
So we want $$u . \mathcal{F}_y(f(t)) = G(y)$$
$$G(y) = \mathcal{F}_y\left(\int\limits_{0}^{t} f(x) dx\right) = \frac{F(y)}{i 2 \pi y} + C \delta(y)$$, where $C = \int\limits_{0}^{\infty} f(x) dx$
Therefore we cannot find a $u$ unless $C=0$. But $\fbox{$\text{if } C = 0, u = \frac{1}{i 2 \pi y}$}$
