Showing $\hat{\tilde{f}}=\tilde{\hat{f}}$ where $\hat{f}$ is the Fourier transform, and $\tilde{f}(x) = f(-x)$

Question

I'm trying to prove that $\hat{\tilde{f}}=\tilde{\hat{f}}$ for any integrable function $f$, where $\hat{f}$ denotes the Fourier transform of $f$ and $\tilde{f}$ denotes the mapping $f(x)\to f(-x)$, though I'm currently stuck at $$\hat{\tilde{f}}(x)=\int_{\mathbb{R}^n}f(-t)e^{-2\pi it\cdot x}dt\tag*{(1)}$$ and $$\tilde{\hat{f}}(x)=\int_{\mathbb{R}^n}f(t)e^{2\pi it\cdot x}dt\tag*{(2)}.$$ Surely the substitution $u=-t$ in $(1)$ gives $$\hat{\tilde{f}}(x)=\int_{\mathbb{R}^n}-f(u)e^{2\pi iu\cdot x}du,$$ which in general is not equal to $(2)$? Many thanks in advance.

Answer 1

You need to be careful when doing the change of variables. The differentials change by the absolute value of the determinant of the Jacobian of the transformation, which in this case is $|(-1)^n| = 1$.

You can see your error if you write your integrals as iterated integrals. Write $\hat{\tilde{f}}(x)$ as $\int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}f(-t)e^{-2\pi it\cdot x}dt_1\dots dt_n$. Then the variable substitution $u = -t$ corresponds to the $n$ variable substitutions $u_k = - t_k$. For each $k$ we see that $du_k = -dt_k$, but also, the limits of integration swap. That is,

\begin{align*} \hat{\tilde{f}}(x) &= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}f(-t)e^{-2\pi it\cdot x}dt_1\dots dt_n\\ &= \int_{\infty}^{-\infty}\dots\int_{\infty}^{-\infty}f(u)e^{2\pi iu\cdot x}(-1)^ndu_1\dots du_n\\ &= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}(-1)^nf(u)e^{2\pi iu\cdot x}(-1)^ndu_1\dots du_n\\ &= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}f(u)e^{2\pi iu\cdot x}|(-1)^n|du_1\dots du_n\\ &= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}f(u)e^{2\pi iu\cdot x}du_1\dots du_n\\ &= \int_{\mathbb{R}^n}f(u)e^{2\pi iu\cdot x}du. \end{align*}

Replacing $u$ by $t$, you then obtain $\hat{\tilde{f}}(x) = \tilde{\hat{f}}(x)$.

Answer 2

If $g\in L^1(\Bbb R^d)$ then $$\int_{\Bbb R^d}g(-t)\,dt=\int_{\Bbb R^d}g(t)\,dt.\quad(*)$$Note the minus sign you think should be there is missing; erase it from what you've done on your problem and you're done.

Of course this raises the question of why the extra minus sign is there in Calc I but not here. Before getting into that, you should prove (*) from the definitions, starting with $g=\chi_E$.

And you should note that an extra minus sign in (*) would make no sense if the integral is the "area under the graph"; the change of variables doesn't change the sign of that area.

And now we begin to suspect that Calc I is wrong. No, the point is just that antiderivatives are not integrals. Say $F'=f$ (in one variable, and ignoring little details like convergence). Then $-F(-t)$ is an antiderivative for $f(-t)$. There's the extra minus sign. But it goes away again when we look at the integral: $$\int_{-\infty}^\infty f(-t)=-\int_\infty^{-\infty}f(t)=\int_{-\infty}^\infty f(t).$$

Which is to say Calc I agrees with (*), it just doesn't look that way because of the emphasis on antiderivatives instead of integrals.