If $f$ is bounded and has bounded derivative on $\mathbb{R}$, then $xf(x)\in L^1\Rightarrow xf(x)\in L^\infty$? Let $f$ be a differentiable and bounded function on $\mathbb{R}$ which has bounded derivative on $\mathbb{R}$. Assume that $\int_{\mathbb{R}}|xf(x)|dx<\infty.$ Does this imply that $xf(x)$ is bounded on $\mathbb{R}$?
I know that in general $g\in L^1$ does not imply $g\in L^\infty$ (e.g., the examples here), but in this case our function is a product of $2$ "good" functions - the function $x$ and a bounded function $f$ which has bounded derivative on $\mathbb{R}$, so probably in this case the implication $xf(x)\in L^1\Rightarrow xf(x)\in L^\infty$ is correct.
 A: What you want to show is not true.
Instead of constructing an explicit counterexample, I will prove its existence.
Let
$$
  X := \{ f \in C_b^1 (\mathbb{R}) : \| x \cdot f \|_{L^1} < \infty \},
$$
equipped with the norm
$$
  \| f\|_X := \|f\|_{L^\infty} + \|f'\|_{L^\infty} + \|x \cdot f \|_{L^1}.
$$
Here, $C_b^1(\mathbb{R})$ denotes the space of all $C^1$ functions that are bounded
with bounded derivative.
It is not difficult to show (do it!) that $X$ is a Banach space.
Now, if your claim was true, the linear operator
$$
  T : \quad
  X \to L^\infty(\mathbb{R}), \quad
  f \mapsto x \cdot f
$$
would be well-defined.
It is not hard to show that $T$ has a closed graph;
indeed, if $f_n \to f$ in $X$ and $T f_n \to g$, then $f_n \to f$ pointwise (in fact uniformly)
and hence $T f_n = x \cdot f_n \to x \cdot f$ pointwise.
Since also $T f_n \to g$ pointwise almost everywhere, this implies $g = x \cdot f = T f$.
Thus, by the closed graph theorem, $T$ is a bounded operator.
We will show below that this is not true, which then yields the desired contradiction.

Fix $h \in C_c^\infty ( (-1,1) )$ with $h(0) = 1$.
Choose $\alpha = 2$ and $\beta = 1$, and for $n \in \mathbb{N}$ define
$$
  f_n : \quad
  \mathbb{R} \to \mathbb{R}, \quad
  f_n (x) = n^{-\beta} \cdot h \big( n \cdot (x - n^\alpha) \big)
  .
$$
Then
$$
  \| f_n ' (x) \|_{L^\infty}
  = \| n^{1 - \beta} \cdot h' \big( n \cdot (x - n^\alpha) \big) \|_{L^\infty}
  \leq n^{1 - \beta} \cdot \| h' \|_{L^\infty}
  =    \| h' \|_{L^\infty}
  .
$$
Likewise, $\| f_n \|_{L^\infty} = n^{-\beta} \| h \|_{L^\infty} \leq \| h \|_{L^\infty}$
for all $n \in \mathbb{N}$.
Finally, note that if $f_n (x) \neq 0$, then $|n \cdot (x - n^\alpha)| \leq 1$ and hence
$|x - n^\alpha| \leq \frac{1}{n}$, so that $|x| \leq n^\alpha + n^{-1} \leq 1 + n^\alpha \leq 2 \, n^\alpha$.
Therefore, it is easy to see
$$
  \| x \cdot f_n \|_{L^1}
  \leq 2 n^\alpha \cdot \| f_n \|_{L^1}
  = 2 n^{\alpha - \beta - 1} \cdot \| h \|_{L^1}
  = 2 \| h \|_{L^1}
  .
$$
In other words, we have shown that $\| f_n \|_X \leq C$ for a constant $C > 0$ independent of $n$.
Since $T$ is bounded, this means that $\| x \cdot f_n \|_{L^\infty} = \| T f_n \|_{L^\infty}$
is bounded as well.
We will now show that this is false; indeed,
$$
  \| x \cdot f_n \|_{L^\infty}
  \geq \frac{n^\alpha}{n^\beta} f_n(n^\alpha)
  =    \frac{n^\alpha}{n^\beta} \cdot h \big( n \cdot (n^\alpha - n^\alpha) \big)
  =    n^{\alpha - \beta}
  =    n
  \xrightarrow[n\to\infty]{} \infty
  .
$$
This completes the proof.
A: Here is a counterexample based in part to the family of functions described by @PhoemueX.
Let $\phi\in \mathcal{C}^{\infty}_{00}(\mathbb{R})$ with $0<\phi<1$ with $\operatorname{supp}(\phi)\subset[0,1]$, and $\phi(1/2)=1$.
Define
$$f(x)=\phi(x)+\sum^\infty_{n=1}\frac{1}{n^2}\phi(n^2(x-n^\alpha)),$$
where $\alpha>2$ is to be determined.
For each $n\in\mathbb{N}$,  $g_n(x)=\frac{1}{n^2}\phi(n^2(x-n^\alpha))$ is supported in $I_n:=\big[n^\alpha,n^\alpha+\tfrac{1}{n^2}\big]$, and the  intervals $I_n$ have pairwise disjoint interiors.  Then,
$$\|f\|_\infty\leq \|\phi\|_\infty<\infty,$$
$f$ is differentiable, $$f'(x)=\phi'(x)+\sum^\infty_{n=1}\phi'(n^2(x-n^\alpha)),$$
and
$$\|f'\|_\infty\leq \|\phi'\|_\infty<\infty$$
On the other hand,
\begin{align}
\int_{\mathbb{R}}xf(x)\,dx &=\int^1_0x\phi(x)\,dx +\sum^\infty_{n=1}\frac{1}{n^2}\int^{n^\alpha+\tfrac{1}{n^2}}_{n^\alpha}x\phi(n^2(x-n^\alpha))\,dx\\
&\leq\int^1_0x\phi(x)\,dx +\sum^\infty_{n=1}\frac{n^\alpha+n^{-2}}{n^4}\int^1_0\phi(x)\,dx<\infty
\end{align}
for $2<\alpha<3$. Fix such an $\alpha$. Notice that for $x_n=n^\alpha+\frac{1}{2n^2}$, $$x_nf(x_n)=\frac{n^\alpha+\tfrac{1}{2n^2}}{n^2}\phi(1)\xrightarrow{n\rightarrow\infty}\infty$$
Hence $xf(x)\notin L_\infty$.
