$f\in C^{\infty}(\mathbb{R}^3)\cap L^1\cap L^{\infty}(\mathbb{R}^3)$ implies $|x|^p\cdot |f|^p\in L^1(\mathbb{R}^3)$ for $1Let $f\in L^1\cap L^p(\mathbb{R}^3)$ for some $p\in (1,\infty)$ and take its mollification $f_{\varepsilon}$. Then by Young's inequality $f_{\varepsilon}\in L^1\cap L^{\infty}(\mathbb{R}^3)$. Does this ensure that $|x|^p|f|^p\in L^1(\mathbb{R}^3)$ for any $p\geq 1$?
Unfortunately the integrability does not give some decay to zero as one can define a function by "gluing" triangles with height and ever decreasing width. This function would satisfy this condition as the series $\sum_{k=0}^{\infty} \frac{k^p}{2^{-k}}<\infty$.
I don't know if there is a more "ugly" function which is smooth and in $L^1\cap L^{\infty}(\mathbb{R}^3)$ so that $|x|^p|f|^p$ is not integrable. Or can one prove this statement? I would be grateful about any answers!
 A: There are bounded counterexamples (your 'triangles') for all $1\le p\le \infty$, and decaying counterexamples for $1\le p < \infty$. Here's the details for the decaying type.
Step 1: one-dimensional problem
Set $\sigma>1$, $\newcommand{\bangle}[1]{\langle#1\rangle}$ $\bangle r := (e+r^2)^{1/2}$, and define $F:\mathbb R \to (0,\infty)$ by
$$F(r) = \frac 1{\bangle r (\log \bangle r)^\sigma}.$$ Then $\int_0^\infty F(r) dr < \infty$, $F$ is smooth on $\mathbb R$, and $F$ decays at infinity. But not a lot. So this is a counterexample to the 1D version of your problem. That is, $F\in L^1\cap L^\infty$ but $rF(r)\notin L^p$ for all $p\in[1,\infty)$.
Step 2: embed correctly in higher dimensions
Let $\phi=\phi(r)$ be any nontrivial, smooth, compactly supported function.
Now, consider $f:\mathbb R^d\to \mathbb R$ defined by $$f(x) := F(x_1) \phi(x_2)\phi(x_3)\dots \phi(x_d).$$
Now observe that

*

*$\|f\|_{L^1(\mathbb R^d)} = \|F\|_{L^1(\mathbb R)}\|\phi\|_{L^1(\mathbb R)}^{d-1}<\infty$.

*$f$ decays at infinity as $F$ does, and $\phi$ is compactly supported. And $f$ is smooth on $\mathbb R^d$. So $f\in L^\infty$.

*Since $|x|\ge |x_1|$, $$\int_{\mathbb R^d} |x|^p |f(x)|^p dx \ge \|\phi\|_{L^p(\mathbb R)}^{(d-1)p} \| x_1 F(x_1)\|_{L^p(\mathbb R)}^p = \infty.$$

In fact, $|r|^\epsilon F \notin L^p$ for all $p\in[1,\infty)$ and all $\epsilon >0$, and the $\epsilon<0$ case is somehow more clear, so essentially no result of 'this type' is true.
For $p=\infty$ I now sketch a variant of the triangle sum function to dispel any doubt that smoothness can help.
Let $\phi$ be any nontrivial smooth  function compactly supported in the ball of radius $1/2$ around the origin. Then $\phi_k(x) :=  \phi(k^{1+\delta}x)$ has $L^1$ norm $\|\phi_k\|_{L^1} = \|\phi\|_{L^1}k^{-1-\delta} \in \ell^1_k$. Now define the sum of disjointly supported functions
$$ g(x) := \sum_{k=1}^\infty \phi_k(x-2^k)$$
clearly $g(x)\in L^1\cap L^\infty\cap C^\infty$, and $x^{\epsilon} g \notin L^\infty$.
In fact, on the support of $\phi_k(x-2^k)$, $x\ge C 2^k$. So
$$ \int_{\mathbb R} |x|^{\epsilon p} |g(x)|^p dx \ge C \sum_{k=1}^\infty 2^{ \epsilon kp} \int_{\mathbb R} |\phi|^p dx=\infty.$$
We used that $\|\phi_k\|_{L^p}$ has at most polynomial decay in $k$.  And then you can also repeat the argument to turn this into a higher dimensional counterexample.
A: No. Take for example $f(x) =\frac{1}{1+x^2}$ for $x \in \mathbb{R}$. Then, $f \in L^1 \cap L^{\infty} \cap C^{\infty}$ (see here for a proof that $f$ is even analytic), however $$\int_{-\infty}^{\infty} \frac{|x|}{1+x^2} dx = 2 \int_{0}^{\infty} \frac{x}{1+x^2}dx  = \infty,$$
since $\frac{x}{1+x^2} \approx 1/x$ for large values of $x$. You can embed this into three dimensions by setting $g(x,y,z) = f(x)f(y)f(z)$ and you will get the analogous result there.
