Convolution of an $L^p$ function with a Compactly supported continuous function. What does it give? CLARIFICATION : In the following discussion $g \in C_c(\Bbb R^n)$ is always assumed to be non-zero.

I know that if for $1 ≤ p ≤ 2$ take a $f ∈ L^ p (\Bbb R^n )$, and  a $g ∈ C_ c (\Bbb R^n )$ such that $f ∗ g ≡ 0$ it is true that $f(x) = 0$, for almost every $x ∈ \Bbb R^n$ . 

I have proved this using the facts that the fourier transform of $g$ can be seen as restriction of an entire function on $\Bbb R$ and then   $f ∗ g ≡ 0 \implies \hat{(f*g)}=0 \implies \hat f . \hat g=0$ would imply from Uniqueness principle that  $\hat f$ must vanish except at most on a set of measure zero, then injectivity of fourier transform yields that $f =0$ a.e.
My question is 

What can be said  if we replace the above range of $p$ to be $(2,\infty]$ ?

Since $L^\infty$ is such a troublesome space to work with in Fourier Analysis, I'd want to treat it in a separate case.
So my questions are : 

$(\mathscr{Q.1})$ If $f \in L^p(\Bbb R^n)$ for some $2< p<\infty$ and $g \in C_c(\Bbb R^n)$ and we have $f*g \equiv 0 $ , then can we say anything analogous to the case when $1 \le p \le 2$ ? 

For $2 < p < \infty$ , I am unable justify the steps $\hat{(f*g)}=0 \implies \hat f . \hat g=0$ and after all, here we are dealing with tempered distributions, so can somehow we use its support ?

$(\mathscr{Q.2})$ If $f \in L^\infty(\Bbb R^n)$ and $g \in C_c(\Bbb R^n)$ and we have $f*g \equiv 0 $ , then can we say anything analogous to the case when $1 \le p \le 2$ ?

 A: WARNING. This answer is correct only in dimension $n=1$.

The answer to Q2 (the $p=\infty$ case) is negative; a counterexample is $f=1$ and $g$ such that $\hat{g}(0)=0$. Indeed,
$$
1\ast g(x)=\int_{\mathbb R^n} g(y)\, dy=\hat{g}(0)=0.$$
The answer to Q1 is affirmative.

EDIT. I just found a mistake. The following is correct only for $n=1$. I still think that the result is true in generality, but it needs some more work.

If $f\ast g=0$, then $\hat{f}\hat{g}=0$; note that this last product makes sense, as $\hat{g}$ is a smooth function, and actually it is analytic, while $\hat{f}$ is a tempered distribution.

Now, $\hat{g}$ only has isolated zeros, by analyticity; denote them by $z_1, z_2, \ldots$.

EDIT. The above needs not be true for $n>1$. For example, the compactly supported function $g(x)=\prod_{j=1}^n \mathbf 1_{[-1, 1]}(x_j)$ has the Fourier transform
$$
\hat{g}(\xi)=\prod_{j=1}^n \frac{\sin \xi_j}{\xi_j}, $$
which is analytic and which has non-isolated zeros. END EDIT.
Thus, $\hat{f}\hat{g}=0$ implies that $\hat{f}$ is a distribution supported in $\{z_1, z_2, \ldots\}$, and the only such distributions have the form
$$
\hat{f}=\sum_{j=1}^\infty \sum_{m\in N(j)} a_{mj} \partial^m \delta_{z_j}, $$
where $N(j)$ is a finite set of multi-indices, and $a_{mj}\in \mathbb C$. $^{[1]}$
Therefore, inverting the Fourier transform, we see that
$$
f=\sum_{j=1}^\infty \sum_{m\in N(j)} a_{mj} i^{|m|} e^{i x\cdot z_j} x^m, $$
which is a contradiction with the assumption $f\in L^p(\mathbb R^n)$, with $p<\infty$.

$^{[1]}$ A multi-index $m$ is a $n$-uple $(m_1, m_2, \ldots, m_n)\in\mathbb N^n$. We denote $\partial^m:=\partial_{x_1}^{m_1}\ldots \partial_{x_n}^{m_n}$ and $x^m :=x_1^{m_1}\ldots x_n^{m_n}$ and $\lvert m\rvert:= m_1+\ldots + m_n$. This is a standard terminology of distribution theory.
A: The result is not true for $n\geq 2$, at least not in the full range: Let $p_0=2n/(n-1)$, then for every $p>p_0$ there exists $f \in L^p(\mathbb{R}^n)$ and $g\in C_c(\mathbb{R}^n)$ such that $g\neq 0 \neq f$ and $f*g=0$.
We'll take $f= \check{\sigma}$, where $\sigma= \mathcal{H}^{n-1}|_{\mathbb{S}^{n-1}}$ is the surface measure on the sphere. It's known (see for example Grafakos's book "Modern Fourier Analysis", it's an appendix if I recall correctly) 
$$
|f(x)|\lesssim (1+|x|)^{-(n-1)/2}.
$$
Therefore $f\in L^p(\mathbb{R}^n)$ for every $p>p_0$. 
Now consider the function $H:\mathbb{C}^n\to \mathbb{C}$ given by 
$$
H(z)=e^{i(z_1+\ldots + z_n)}(z_1^2+\ldots +z_n^2-1),
$$
where $z=(z_1,\ldots,z_n)$. Then $H$ is entire and 
$$
|H(z)|\leq (1+|z|)^2 e^{|\text{Im}(z)|},
$$
so that by a Paley-Wiener Theorem (see Strichartz's book on distributions, or the Wikipedia page) $H=\hat{h}$ for a compactly supported distribution $h$. Now take $g=h*\eta$ for some $\eta\in C_c^\infty(\mathbb{R}^n)$ to obtain that $g\in C_c(\mathbb{R}^n)$ and $\hat{g}= H \hat{\eta}$, which vanishes on the unit sphere.
Note that for $n=1$ we have no decay since the measure $\sigma$ is just a sum of Dirac masses. This agrees with @GiuseppeNegro's result that $p=\infty$ is the only failure there. On the other hand for $p>2$ and $n\geq 2$ Fourier transforms of $L^p$ functions may be supported on lower dimensional sets. I'm not sure that $p_0$ is sharp though, so one question remains:
Is the original question true for $p\in (2,2n/(n-1)]$?
A: I submit a solution to the remaining open question, which I repeat here to ease reading.

Question (Q2). Let $2<p<\infty$. Are there $0\ne f\in L^p(\mathbb R^n)$ and $0\ne g\in C_c(\mathbb R^n)$ such that  $f\ast g=0$?
Answer.
They exist if and only if $p>\frac{2n}{n-1}$.

Proof (SKETCH - INCOMPLETE). The "if" part has been cleverly solved by Jose27. We turn to the "only if". The equation $f\ast g=0$ is equivalent to $\hat{f}\hat{g}=0$, and $\hat{g}$ is analytic. Thus, there is a solution if and only if there is $f\in L^p(\mathbb R^n)$ such that $\hat{f}$ is supported on the zero set of $\hat{g}$.
We assume that the zero set of $\hat{g}$ is the unit sphere. This means that $\hat{f}$ is supported on the unit sphere, hence that it is of the form $\hat{f}=F\, d\sigma$, where $d\sigma$ denotes the surface measure. Now, the principle of stationary phase implies that
$$
f(x)=\int_{\mathbb S^{n-1}} F(\xi)e^{ix\cdot \xi}\, d\sigma(\xi) \sim \frac{C}{\lvert x \rvert^{\frac{n-1}{2}}},\quad |x|\to \infty, $$
where $\sim$ means "asymptotically equivalent" (i.e., the ratio tends to 1). And so, if $A>0$ is sufficiently big,
$$
\int_{|x|>A}|f(x)|^p\, dx \ge C\int_{|x|>A} |x|^{-p\frac{n-1}{2}}\, dx, $$
which is infinite if $p\le \frac{2n}{n-1}$.
TO DO.

*

*Explain why it is sufficient to assume that $\hat{g}=0$ is the sphere. The hand-waving reason is that the zero set must have nonvanishing Gaussian curvature, otherwise $f$ would not decay in one direction and thus fail to be in $L^p$. Once there is nonvanishing Gaussian curvature, we can assume that the surface is either the sphere or the paraboloid, all the others are essentially the same as those two.

*Fill in the details of the principle of stationary phase.


PREVIOUS VERSION OF THIS POST. What follows is obsolete.
Connection with Stein-Tomas extension estimate.
Jose27 unveiled the connection between Q2 and the restriction/extension problem of harmonic analysis.$^{[1]}$  For the case of the sphere, Tomas and Stein proved the following extension inequality
$$
\lVert \widehat{F\, d\sigma}\rVert_{L^{p}(\mathbb R^n)}\le C\lVert F\rVert_{L^2(d\sigma)}, $$
where $C>0$, $\sigma$ denotes the surface measure on the sphere $\mathbb S^{n-1}$, and $$p\ge p_0:=2(n+1)/(n-1).$$ This is related with Q2, because as Jose27 points out, we can take
$$
f=\widehat{F\, d\sigma}, $$
(in Jose's post, $F=1$). We thus get $f\in L^p(\mathbb R^n)$, and now we can construct $g\in C^\infty_c(\mathbb R^n)$ such that $\hat{g}$ vanishes on $\mathbb S^{n-1}$ using the Schwartz-Paley-Wiener theorem. Therefore, $f\ast g$ has a vanishing Fourier transform; that is, $f\ast g=0$.
This implies that the answer to Q2 is negative for $p\ge p_0$.
Can we go below $p_0$? For that, we should find a function $F\in L^2(d\sigma)$ such that $\widehat{Fd\sigma}\in L^p(\mathbb R^n)$ for some $p<p_0$.

The measure $d\sigma$ needs not be the surface measure on the sphere; other surfaces, perhaps of higher codimension, or even curves, should do. Can someone check? The book on harmonic analyis of Stein is the obvious place to start.


[1] "Restriction" and "extension" refer to estimates which are dual to each other. Thus, one can formulate the problem in terms of one or another, but the content is essentially the same. Here, we need the "extension" estimate, but the problem is more often known as "restriction".
