# Sequence of complex $L_1$ functions with $L_1$-divergent limit of the fourier operator

In a task I should first show that $\mathcal{F}(e^{-\frac{x^2}{2 k}})=\sqrt{k} e^{-\frac{k \xi ^2}{2}}$ if $\text{Re}(k)>0$.

Then they say that one may conclude that there is a sequence $(f_k)_{k \in \mathbb{N}}$ such that $\text{ess sup}(|f_j|)=1$, $f_j \in C^\infty$ and $f_j \in L_1$ such that the $L_1$ norm of $\mathcal{F}(f_k)$ diverges to $\infty$, namely $\lim_{k \rightarrow \infty}||\mathcal{F}f_k||_1=\infty$.

My question is how to find such $f_k$. (We consider complex valued functions in the whole task). I could show the identity but I cannot understand at all how we could construct such $f_k$'s as

$$||\mathcal{F}(e^{-\frac{x^2}{2 k}})||_1=\int_{-\infty }^{\infty } \sqrt{k} e^{-\frac{1}{2} \left(k \xi ^2\right)} d\xi=2\pi$$

Therefore you cannot choose the $k$'s wisely to get such a sequence. Can one consider a convolution with a different function to get the result? The problem is for all examples I looked at the property $\text{ess sup}(|f_j|)=1$ was not fulfilled, I hope that I am not overlooking something trivial.

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Take any function $f\in L^1\cap L^\infty$ such that $\|f\|_\infty=1$ and $\hat f\not\in L^1$. For instance, $f$ could be the characteristic function of a bounded interval. Let $$G_j(x)=\sqrt{j\strut\,}\,e^{-\tfrac{x^2}{2j}}\quad\text{and}\quad f_j=G_j\star f.$$ Then $\|f_j\|_\infty\le \|G_j\|_1\|f\|_\infty=1$ and $\hat{f_j}(\xi)=e^{-\tfrac{j\,\xi^2}{2}}\hat f(\xi)$. I leave to you to check that $\|\hat f_j\|_1$ converges to $\infty$ as $j\to0$.