Fourier Transform of 1-Lebesgue Space Function is Bounded

Theorem

Let $n \in \N_{>0}$.

Let $\map {L^1} {\R^n}$ be the complex-valued Lebesgue $1$-space with respect to the Lebesgue measure.

Let $f \in \map {L^1} {\R^n}$.

Let $\hat f$ be the Fourier transform of $f$.

Then:

$\ds \sup_{\mathbf s \mathop \in \R^n} \cmod {\map {\hat f} {\mathbf s} } \le \norm f_{\map {L^1} {\R^n} }$

Proof

For each $\mathbf s \in \R^n$:

\(\ds \cmod { \map {\hat f} {\mathbf s} }\)

\(=\)

\(\ds \cmod { \int_{\R^n} \map f {\mathbf x} e^{-2 \pi i \mathbf x \cdot \mathbf s} \rd \mathbf x }\)

Definition of Fourier Transform

\(\ds \)

\(\le\)

\(\ds \int_{\R^n} \cmod { \map f {\mathbf x} e^{-2 \pi i \mathbf x \cdot \mathbf s} } \rd \mathbf x\)

Triangle Inequality for Integrals

\(\ds \)

\(=\)

\(\ds \int_{\R^n} \cmod { \map f {\mathbf x} } \rd \mathbf x\)

Modulus and Argument of Complex Exponential

\(\ds \)

\(=\)

\(\ds \norm f_{\map {L^1} {\R^n} }\)

Definition of $L^1$ Norm

$\blacksquare$