I was reading the notes "Introduction to Complex Analysis" by Michael E. Taylor, and I am stuck on the following exercise about the Fourier Transform and the space $\mathcal{A}(\mathbb{R})$ defined as follows:
$$\mathcal{A}\left(\mathbb{R}\right):= \left\{ f \in L^{1}(\mathbb{R}) \cap C(\mathbb{R}) : \ f \ \text{bounded} \quad \hat{f} \in L^{1}(\mathbb{R}) \right\}.$$
Exercise 10, chapter 14:
Sharpen the result of Exercise $1$ as follows, using the reasoning in the proof of Proposition 14.6. Assume $ f : \mathbb{R} \rightarrow \mathbb{C} $ is a $ C^1 $ function satisfying $$ |f^{(j)}(x)| \leq C \cdot (1 + |x|)^{-2}, \quad j \leq 1. \tag{14.45A} $$ Then show that $ f \in \mathcal{A}(\mathbb{R}) $. More generally, show that $ f \in \mathcal{A}(\mathbb{R}) $ provided that $ f $ is Lipschitz continuous on $ \mathbb{R} $, $ C^1 $ on $ (-\infty, 0] $ and on $ [0, \infty) $, and $|f^{(j)}(x)| \leq C(1 + |x|)^{-2}$ holds for all $ x \neq 0 $.
Maybe, I should use the following theorem:
Proposition 14.7. Let $f_\nu \in \mathcal{A}(\mathbb{R})$ and $f \in C(\mathbb{R}) \cap L^1(\mathbb{R})$. Assume $f_{\nu} \to f$ in $L^1$-norm, and $\| \hat{f}_{\nu} \|_1 \leq A$ for every $\nu.$ Then $\hat{f} \in L^1(\mathbb{R}),$ i.e, $f \in \mathcal{A}(\mathbb{R})$.
Could you help me at least with the case $C^1$ and give me a hint in the general case? I made an attempt, but I cannot conclude. First, I observed that I made an attempt, but I cannot conclude. First, I observed that $$ \vert \hat{f}(\xi) \vert \leq \int_{\mathbb{R}} \frac{1}{(1+ \vert x\vert)^{2}} \, dx = 2. $$ Consequently, we have that $$ \vert \hat{f}(\xi) \cdot \exp(-2\cdot \epsilon \xi^2) \vert \leq 2, $$ i.e, the Fourier Transform of $f_{\epsilon}(x):= f*g_{\epsilon}(x)$ is limited by a constant indipendent from $\epsilon.$ If I could prove that $f*g_{\epsilon}$ is in $\mathcal{A}(\mathbb{R})$ then I would have concluded because $f*g_{\epsilon}$ approaches $f$ in $L^1.$
Not an answer, but too long for a comment.
You should give the conventions for the Fourier transform and write what $g_\epsilon$ is. Something like $(1/\epsilon) \exp(-x^2/\epsilon)$ I presume?
The difficulty in your approach is to show that the family of function $\hat{f}\hat{g_\epsilon}$ is bounded in $L^1$. Proving that they are bounded i $L^\infty$ is not sufficient.
The other points are true since the function $g_\epsilon$ is in $L^1$ and in $L^\infty$. Since $f$ is also in $L^1$, $f*g_\epsilon$ is in $L^1$ (the convolution of two functions in $L^1$ is still in $L^1$). And any convolution $L^1*L^\infty$ brings a bounded continuous function, so $f*g_\epsilon$ is continuous (and bounded).