Spring 2015 - Problem 4

Solution.

"$\implies$"

Suppose $\hat{f}\p{n} = 0$ for some $n$. Observe that for any linear combination $h\p{x} = \sum_{n=1}^N c_j f\p{x - a_j}$,

Hence, linear combinations of translates of $f$ cannot possibly be dense. Indeed, consider $g\p{x} = e^{inx}$ has $\hat{g}\p{n} = 1$. Because the Fourier transform is a continuous function from $L^1 \to C_0\p{\R}$, if $\set{f_k}_k$ is a sequence of linear combinations of translates of $f$ such that $f_k \to g$ in $L^1$, then we would have$\hat{f_k} \to \hat{g}$ uniformly. However, this is impossible, since $\abs{\hat{f_k}\p{n} - \hat{g}\p{n}} = 1$ for all $k \geq 1$.

"$\impliedby$"

Let $M = \cl{\span\,\set{f\p{x - a}}_{a \in \R}}$. Suppose that $M \neq L^1\p{\p{0, 2\pi}}$ and so there exists $x_0 \in L^1\p{\p{0, 2\pi}} \setminus M$. By Hahn-Banach, there exists a linear functional $L_g \in \p{L^1\p{\p{0, 2\pi}}}^* \simeq L^\infty\p{\p{0, 2\pi}}$, i.e., integration against $g$, such that $L_g\p{x_0} \neq 0$ but $\res{L_g}{M} = 0$. Hence,

for all $a \in \R$ by construction. Notice that by Hölder's inequality, $e^{-i\xi x}fg \in L^1$. Hence, we may apply Fubini's theorem in the following:

But $h = 0$ almost everywhere, so $\hat{h} = 0$ everywhere. Thus,

for all $n$, since $\hat{f}\p{n} \neq 0$. Since $L^\infty\p{\p{0, 2\pi}} \subseteq L^2\p{\p{0, 2\pi}}$, we see that

which is impossible as $g$ was non-zero. Hence, $M = L^1\p{\p{0, 2\pi}}$ to begin with.