Non-Surjectivity of the Fourier Transform

In this article, we wish to prove the following theorem:

Theorem: The Fourier transform from $L^1(\mathbb{R})$ to $\mathcal{C}_0(\mathbb{R})$ is not surjective.

Before that, we will prove some basic facts.

Some Lemmas

Lemma 1: For any $a\ge b \ge 0,$ we have $\left|\int_a^b\frac{\sin(x)}{x}\right|\le 8.$

Proof: If $a>1,$ we have $$\left|\int_{0}^a\frac{\sin(x)}{x}dx\right|\le\left|\int_{0}^1\frac{\sin(x)}{x}dx\right|+\left|\int_{1}^a\frac{\sin(x)}{x}dx\right|$$

$$\le \int_{0}^1\left|\frac{\sin(x)}{x}\right|dx+\left|\frac{-\cos(a)}{a}+\frac{\cos(1)}{1}-\int_{1}^a\frac{\cos(x)}{x^2}dx\right|$$$$\le\int_0^1 dx+\left|\frac{\cos(a)}{a}\right|+\left|\frac{\cos(1)}{1}\right|+\int_1^a\left|\frac{\cos(x)}{x^2}\right|dx\le 1+\frac{1}{a}+1+\int_1^\infty\frac{dx}{x^2}\le 4$$

since $a>1$ and $\int_1^\infty x^{-2}dx=1.$ We also used the fact that $|\sin(x)|\le |x|$ for all $x\in\mathbb{R}.$

If $a\le 1$ then using then $$\left|\int_{0}^a\frac{\sin(x)}{x}dx\right|\le \int_{0}^a\left|\frac{\sin(x)}{x}\right|dx\le \int_0^adx=a\le 1\le 4$$

Hence, for any $a\ge b\ge 0,$ we have $$\left|\int_{b}^a\frac{\sin(x)}{x}\right|=\left|\int_{0}^a\frac{\sin(x)}{x}-\int_{0}^b\frac{\sin(x)}{x}\right|\le \left|\int_{0}^a\frac{\sin(x)}{x}\right|+\left|\int_{0}^b\frac{\sin(x)}{x}\right|\le 8$$

Lemma 2: If $\hat{f}$ is odd for some $f\in L^1(\mathbb R)$ then $f$ is odd a.e.

Proof: Let $h:\mathbb{R}\to\mathbb{R}$ be defined as $h:x\mapsto f(x)+f(-x).$ We have $h\in L^1$, since $$\int |h|dm\le \int(|f(x)|+|f(-x)|)dx\le \int|f(x)|dx+\int|f(-x)|dx=2||f||_{L^1}<\infty$$

Note that for any $\xi\in\mathbb{R},$ we have $$\hat{\tilde{f}}(\xi)=\int f(-x)e^{-2\pi i\xi x}dx=\int f(y)e^{2\pi i\xi y}dx=\hat{f}(-\xi)=-\hat{f}(\xi)$$

Hence, $\hat{h}=\hat{f}+\hat{\tilde{f}}=\hat{f}-\hat{f}=0.$ Hence, $\hat{h}\in L^1$ as well. Consider the function $h_0$ defined as $h_0(x)=\int \hat{h}(\xi)e^{2\pi i \xi x}d\xi=\int 0e^{2\pi i \xi x}d\xi=0.$ Then by the inversion theorem, $h=h_0=0$ a.e. So $f$ is odd a.e.

Proof of the Theorem

Let $g$ be an odd function such that $g=\hat{f}$ for some $f\in L^1.$ Then, $f$ is also odd by Lemma 2. So, for any $\xi\ge 1$ we have $$g(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i \xi x}dx=\int_{-\infty}^0 f(x)e^{-2\pi i \xi x}dx+\int_{0}^\infty f(x)e^{-2\pi i \xi x}dx$$

$$=\int_{0}^\infty f(-y)e^{2\pi i \xi y}dy+\int_{0}^\infty f(x)e^{-2\pi i \xi x}dx=-\int_{0}^\infty f(y)e^{2\pi i \xi y}dy+\int_{0}^\infty f(x)e^{-2\pi i \xi x}dx$$

$$=\int_0^\infty f(x)\big[e^{-2\pi i\xi x}-e^{2\pi i \xi x}\big]dx=-2i\int_0^\infty f(x)\sin(2\pi\xi x)dx$$

Hence, for any $R>1,$ we have $$\int_1^R\frac{g(\xi)}{\xi}d\xi=-2i\int_1^R \int_0^\infty f(x)\frac{\sin(2\pi\xi x)}{\xi} dxd\xi$$

Note that $(x,\xi)\mapsto f(x)\frac{\sin(2\pi\xi x)}{\xi}\chi_{(0,\infty)}(x){\xi}\chi_{[1,R]}(\xi)$ is integrable and using Tonelli's theorem:

$$\int_{(0,\infty)\times[1,R]} \left|f(x)\frac{\sin(2\pi\xi x)}{\xi}\right|d(x,\xi)=\int_0^\infty\int_1^R \left|f(x)\frac{\sin(2\pi\xi x)}{\xi}\right|d\xi dx=\int_0^\infty |f(x)|\int_1^R\left|\frac{\sin(2\pi\xi x)}{\xi}\right|d\xi dx$$

$$\le \int_0^\infty |f(x)|\left|\int_{2\pi x}^{2R\pi x}\frac{\sin(t)}{t}\right| dt dx\le 8\int_0^\infty |f(x)| dx=8||f||_{L^1}<\infty$$

Here, we used Lemma 1 for the bound. Hence, $\left|\int_1^R\frac{g(\xi)}{\xi}d\xi\right|\le 16||f||_{L_1}.$

Since this holds for all $R>1,$ we have $$\sup_{R>1}\left|\int_1^R\frac{g(\xi)}{\xi}d\xi\right|\le 16||f||_{L^1}<\infty\text{ ($\star$)}$$

Now assume, for a contradiction, that the Fourier transform is surjective. Define $g$ as $g(\xi)=\xi$ if $\xi\in [-1,1]$ and $g(\xi)=\frac{\text{sgn}(\xi)}{\ln(e|\xi|)}$ for $\xi\in[-1,1]^c.$ Obviously $g\in \mathcal{C}_0(\mathbb{R}).$ So, by our assumption, $g=\hat{f}$ for some $f\in L^1.$

Further, $g$ is odd. Hence, by our above argument ($\star$), $$\sup_{R>1}\int_1^R\frac{g(\xi)}{\xi}d\xi=\sup_{R>1}\int_1^R\frac{1}{\xi\ln(e\xi)}d\xi<\infty$$

However, this is a contradiction (by the integral test, applicable as $g$ is continuous, positive and decreasing in $[1,\infty),$ and Theorem 3.29 with $p=1$ (Baby Rudin)). So $g$ cannot be the fourier transform of an $L^1$ function and hence the Fourier transform is not surjective (with respect to co-domain $\mathcal{C}_0(\mathbb{R})$).