Find $f\in L^1(\mathbb{R})$ such that $\|\int_{-N}^N \hat{f}(\xi)e^{2\pi i\xi \cdot}\operatorname{d}\xi-f\|_1\nrightarrow 0, N\to\infty?$ I'm wondering how to find a function $f\in L^1(\mathbb{R})$ such that, if $\hat{f}$ is its Fourier transform, i.e.:
$$\forall \xi\in\mathbb{R},\hat f (\xi):=\int_\mathbb{R} f(t)e^{-2\pi i\xi t}\operatorname{d}t$$
then
$$\left\|\int_{-N}^N \hat{f}(\xi)e^{2\pi i\xi \cdot}\operatorname{d}\xi-f\right\|_1=\int_{\mathbb{R}}\left|\int_{-N}^N \hat{f}(\xi)e^{2\pi i\xi x}\operatorname{d}\xi-f(x)\right|\operatorname{d}x\nrightarrow 0, N\to\infty$$
I know that such a function must exist since in the book Fourier Analysis by Javier Duoandikoetxea on page 59 it is stated without proof that for $f\in L^1(\mathbb{R})$, in general we can't expect that $\int_{-N}^N \hat{f}(\xi)e^{2\pi i\xi \cdot}\operatorname{d}\xi$ converges in $L^1$ norm to $f$ but that anyway it converges to $f$ in measure.
Can anyone provide any reference or just a proof that guarantees the existence of such a function?
 A: Let $f(t)=\begin{cases}1&-\frac12\le t\le \frac12\\0&\text{otherwise}\end{cases}$.
Then
$$\hat{f}(s) = \int_{-\frac12}^{\frac12} e^{-2\pi ist}\,dt = \frac1{-2\pi is}\left(e^{-\pi is}-e^{\pi is}\right)=\frac{\sin(\pi s)}{\pi s}$$
Integrate that against $e^{2\pi isx}$ on $[-N,N]$ and we get
\begin{align*}g_N(x) &= \int_{-N}^N \frac{\sin(\pi s)}{\pi s}e^{2\pi isx}\,ds = 2\int_{0}^N \frac{\sin(\pi s)}{\pi s}\cos(2\pi sx)\,ds\\
g_N'(x) &= -2\int_{0}^N 2\sin(\pi s)\sin(2\pi sx)\,ds = 2\int_0^N \cos(\pi s(1+2x))-\cos(\pi s(1-2x))\,ds\\
g_N'(x) &= \frac{2\sin(\pi N(2x+1))}{\pi(2x+1)} - \frac{2\sin(\pi N(2x-1))}{\pi(2x-1)}\end{align*}
For some $N$, that's pretty small. The two $\sin$ terms are equal for integer $N$, leaving us with something that's $O(x^{-2})$ as $x\to\infty$. For other $N$, it grows larger. If $N-\frac12$ is an integer, then $\sin(2\pi Nx+\pi N)=-\sin(2\pi Nx-\pi N)$ and
$$g_N'(x) = \frac{8x\sin(2\pi Nx+\pi N)}{(4x^2-1)\pi} = \frac{\pm 8x\cos(2\pi Nx)}{(4x^2-1)\pi}$$
For large $x$ and half-integer $N$, then, $g_N$ changes by
$$\int_{(k-\frac12)/N}^{(k+\frac12)/N}g_N'(x)\,dx\approx \int_{(k-\frac12)/N}^{(k+\frac12)/N}\frac{\pm 8k/N\cos(2\pi Nx)}{4k^2\pi/N^2}\,dx = \frac{\pm 2}{k\pi^2}\approx\frac{\pm 2}{\pi^2 Nx}$$
between extremes, as $x$ changes by $\frac1N$. That's too much variation for a $L^1$ function; $|g_N(x)|$ will be at least $\frac cx$ for some fixed $c>0$ most of the time, and thus won't be in $L^1$.
So then, even in this simple example, we don't get $L^1$ convergence - a sequence of functions that aren't in $L^1$ at all can't possibly converge to anything in $L^1$. This was the first example I tried - just pick $f$ far enough from smoothness that $\hat{f}$ isn't $L^1$, and the failure is practically inevitable.
