Fourier transform identity. $\int\limits_{-\infty}^{\infty}\hat f(x)g(x)\,dx=\int\limits_{-\infty}^{\infty} f(x)\hat g(x)\,dx$

Suppose $f,g\in L^1(\Bbb R)$

I'm trying to show that $$\int\limits_{-\infty}^{\infty}\hat f(x)g(x)\,dx=\int\limits_{-\infty}^{\infty} f(x)\hat g(x)\,dx$$ where $\hat f(x)=\int\limits_{-\infty}^{+\infty}f(y)e^{-2\pi ixy}\,dy$ is the Fourier transform of $f$.

So far I have

$$\int\limits_{-\infty}^{\infty}\hat f(x)g(x)\,dx= \int\limits_{-\infty}^{\infty} \left(\int\limits_{-\infty}^{+\infty} f(y)e^{-2\pi ixy}\,dy\right)g(x)\,dx= \int\limits_{-\infty}^{\infty} \left(\int\limits_{-\infty}^{+\infty} f(y)g(x)e^{-2\pi ixy}\,dy\right)\,dx$$

and by using Fubini theorem (here I am not sure how well written this is) we can change the order of integration:

$$=\int\int f(y)g(x)e^{-2\pi ixy}\,dx\,dy=\int \int g(x)e^{-2\pi ixy} f(y) \, dx \, dy = \int\hat g(y)f(y)\,dy$$

• You are missing a $dx$ in the second to last term and I guess it would be good to give an argument for the application of Fubini's theorem. ($f,g \in L^1$ and the exponential function is bounded in absolute value by 1). Apart from that, I think the proof is fine.
– pcp
Commented May 3, 2018 at 12:26
• What goes hand in hand with the explanation of why Fubini can be applied, is an argument, why both integrals at the beginning exist. You should start with that.
– pcp
Commented May 3, 2018 at 12:31
• Where did you find this identity stated? Commented May 3, 2018 at 18:00
• It was one of my exercises Commented May 3, 2018 at 18:01

Your argument largely looks ok. Let's follow up on P.Pet's comment. Consider $$\iint\limits_{\mathbb R^2} \left| f(x)g(y) e^{-2\pi ixy} \right| \, d(x,y).$$ Here I wrote $d(x,y)$ and not $dx\,dy$ or $dy\,dx.$ This is an integral with respect to Lebesgue measure in the plane. Since $|e^{-2\pi ixy}| = 1,$ this is equal to $$\iint\limits_{\mathbb R^2} \left| f(x) g(y) \right| \, d(x,y).$$ Tonelli's theorem says the double integral of a non-negative function is equal to either of the two iterated integrals, regardless of whether its value is finite or infinite. Thus we have \begin{align} & \iint\limits_{\mathbb R^2} \left| f(x)g(y) e^{-2\pi ixy} \right| \, d(x,y) \\[10pt] = {} & \iint\limits_{\mathbb R^2} \left| f(x) g(y) \right| \, d(x,y) \\[10pt] = {} & \int\limits_{\mathbb R} \left( \int\limits_{\mathbb R} |f(x)g(y)| \, dx \right) dy & & \text{by Tonelli's theorem} \\[10pt] = {} & \int\limits_{\mathbb R} \left( |g(y)| \int\limits_{\mathbb R} |f(x)|\,dx \right) dy & & \text{(pulling a constant out of the inner integral)} \\[10pt] = {} & \int\limits_{\mathbb R} |g(y)|\, dy \cdot \int\limits_{\mathbb R} |f(x)|\,dx & & \text{(pulling a constant out of the outer integral)} \\[10pt] < {} & {+\infty} & & \text{by a hypothesis stated in the posted question.} \end{align} The fact that the integral of the absolute value with respect to the $2$-dimensional measure is finite is what tells you Fubini's theorem may be applied.