Supremum of absolute value of the Fourier transform equals $1$, and it is attained exactly at $0$

Question

Suppose that $f \in L^1(\mathbb{R}^n)$, $f \ge 0$, $\|f\|_{L^1} = 1$. How do I see that $\sup_{\xi\in\mathbb{R}^n} |\mathcal{F}(f)(\xi)| = 1$, and it is attained exactly at $0$?

Answer 1

You should define all your terms. I presume $\cal F$ is the Fourier transform. The standard formula is that $$\cal F f(x) = \int_{\mathbb R} e^{-2\pi i x y} f(y) \, dy.$$ Since $2\pi i x y$ is purely imaginary or zero, $|e^{-2\pi i xy}| = 1$. Apply the triangle inequality to get $$|\cal F f(x)| \le \int_{\mathbb R} |e^{-2\pi i x y} f(y)| \, dy = \int_{\mathbb R} |f(y)| \, dy = 1$$ for any $x$. Since $e^{0} = 1$ you get moreover $$\cal F f(0) = \int_{\mathbb R} f(y) \, dy = \int_{\mathbb R} |f(y)| \, dy = 1$$ since $f \ge 0$.

Answer 2

First, note that we have$$|\mathcal{F}f(\xi)| = \left|\int_{\mathbb{R}^d} f(x)e^{-ix \cdot \xi}dx\right| \le \int_{\mathbb{R}^d} |f(x)|\,dx = 1 = \mathcal{F}f(0),$$where the final equality comes from the fact that $f \ge 0$. Since $f$ is real-valued, we may decompose $\mathcal{F}f(\xi)$ into its real and imaginary parts as$$\mathcal{F}f(\xi) = \int_{\mathbb{R}^d} f(x)\cos(x \cdot \xi)\,dx - i \int_{\mathbb{R}^d} f(x)\sin(x \cdot \xi)\,dx.$$Now, observe that since $f(x)\cos(x \cdot \xi) \le f(x)$, the real part of the Fourier transform satisfies$$\text{Re}\,\mathcal{F}f(\xi) = \int_{\mathbb{R}^d} f(x)\cos(x \cdot \xi)\,dx \le \int_{\mathbb{R}^d} f(x)\,dx.\tag*{$(*)$}$$If $\xi \neq 0$, $\cos(x \cdot \xi) < 1$ away from a set of measure zero, as this set is easily seen to be a countable union of hyperplanes orthogonal to the line determined by $\xi$, and we can thus have equality in $(*)$ only if $f = 0$ almost everywhere away from this set. However, this obviously would imply $\|f\|_{L^1} = 0$, contradicting our assumption, so we conclude that the inequality in $(*)$ is strict whenever $\xi \neq 0$.

To prove $0$ is the only point where $|\mathcal{F}(f)|$ attains $1$, we argue by contradiction. Suppose $\xi \neq 0$ satisfies $|\mathcal{F}(f)(\xi)| = 1$. Notice we already know $\text{Re}\,\mathcal{F}f(\xi) < 1$. $\mathcal{F}(f)$ is not a positive real number. Therefore, by a complex rotation, we can make it $1$, that is, $\mathcal{F}(f)(\xi)e^{-i\theta} = 1$ for some $\theta \in (0, 2\pi)$. However, for any $\theta \in (0, 2\pi)$,$$\text{Re}\,\mathcal{F}(f)(\xi)e^{-i\theta} = \int_{\mathbb{R}^d}f(x)\cos(x \cdot \xi + \theta)\,dx \le \int_{\mathbb{R}^d} f(x)\,dx = 1.$$For the same reason as before, the inequality is strict if $\xi \neq 0$. Contradiction.