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I want to prove it through the hint given in the notes available online(link provided below). It says first prove that if $f\in C_c^2(\mathbb{R})$, then $\hat{f}\in L_1(\mathbb{R})$; and hence conclude range of fourier transform on $L_1(\mathbb{R})$ is dense in $C_0(\mathbb{R})$. Conclusion part is okay, but I am unable to prove that if $f\in C_c^2(\mathbb{R})$, then $\hat{f}\in L_1(\mathbb{R})$. For those who are not familiar with the terminology, let me recap: $C_c^2(\mathbb{R})$ denotes the collection of continuous functions with compact support which are square integrable.

Link of the notes: http://people.math.gatech.edu/~heil/handouts/chap1.pdf

see page $\textbf{72}$, Exercise $\textbf{1.100}$

Rakesh Balhara
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    Integrate by parts twice. – zhw. Apr 14 '15 at 06:36
  • sorry i misunderstood the terminology, here $C_c^2(\mathbb{R})$ denotes the collection of continuous functions with compact support which are twice continuously differentiable. Now the question is over. :) @zhw thanks for your comment which made me rethink about the terminology. – Rakesh Balhara Apr 14 '15 at 06:44
  • I tried to write out an answer to this question, and stumbled upon a mistake I couldn't go around. OP or @zhw. can you share your proof? – Silvia Ghinassi Aug 16 '15 at 16:05

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For $x\ne 0,$

$$\hat f (x) =\int_\infty^\infty f(t)e^{-ixt}\, dt =1/(ix)\int_\infty^\infty f'(t)e^{-ixt}\, dt = 1/(ix)^2\int_\infty^\infty f''(t)e^{-ixt}\, dt.$$

I've integrated by parts twice; each time the boundary terms disappear because of compact support. The last expression is $1/x^2$ times a function in $C_0.$ We already know $\hat f$ is bounded. The above shows it is bounded by a constant times $1/x^2$ for $|x|$ large. Hence $\hat f \in L^1.$

zhw.
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