# Integrable function whose Fourier transform is constant

Does there exist an integrable function whose continuous Fourier transform equals a constant $c \neq 0$ on $\mathbb{R}$? That is, does there exist a function $f \in L_1$ such that $\hat{f}(x) = \int_\mathbb{R} f(t)e^{-2\pi i xt}dt = c$, $\forall x \in \mathbb{R}$, with $c \neq 0$?

I don't think so, but I'm not really sure how I would go about proving/disproving this feeling. I have not yet been introduced to Fourier transforms. A hint would be appreciated!

• Dirac Delta Function? – QuantumFool Oct 25 '16 at 4:31
• I'm not a mathematician, but the delta function has a constant Fourier transform. This has important consequences in physics, such as having a localized position in QM leads to having a completely unknown momentum! – Doug Oct 25 '16 at 4:31
• I'm not very advanced in math and I've heard that that probably counts as a distribution not a function, but anyways it's useful in Physics. – QuantumFool Oct 25 '16 at 4:32
• @QuantumFool "Function" – Omnomnomnom Oct 25 '16 at 4:32
• @Doug haha you beat me by ~5 seconds. I'm a physics person so that's where I learned about Dirac Deltas and all – QuantumFool Oct 25 '16 at 4:33

Suppose there exists an integrable function $f$ such that \begin{align} \int^\infty_{-\infty} f(x) e^{-2\pi i x \xi}\ dx = 1 \end{align} then it follows \begin{align} \int^{\infty}_{-\infty} f(x)e^{-2\pi i x n} \ dx = 1 \end{align} for all $n$ but this contradictions Riemann Lebesgue lemma.
• Why do you even mention the integer values $n$? – zhw. Oct 25 '16 at 4:48
• @zhw I was thinking about Fourier series when I wrote it. You are right that I don't need to think about $n$. – Jacky Chong Oct 25 '16 at 4:50
It is impossible to evaluate that integral outright, since it doesn't converge. That being said, here's a reasonable approach: it is known that for any functions $f$ and $g$, we have $$\widehat{fg}(x) = (\hat f*\hat g)(x) = \int_{-\infty}^\infty \hat f(y)\hat g(x-y)\,dy$$ Thus, if this property is to be maintained, our $\hat f$ should be such that for any $g$: $$\hat g(x) = \widehat{fg}(x) = \int_{-\infty}^\infty \hat f(y)\hat g(x-y)\,dy$$ that is, $\hat f$ must have the sifting property. In fact, this tells us that $\hat f$ must be the dirac delta "function" (which is, strictly speaking, not a function).