4. If the solution set of the inequality $k x^{2}-2|x-1|+$ $6 k<0$ with respect to $x$ is an empty set, then the range of values for $k$ is $\qquad$.
4. $k \geqslant \frac{1+\sqrt{7}}{6}$.

From the problem, for every $x \in \mathbf{R}$, we have
$$
k x^{2}-2|x-1|+6 k \geqslant 0,
$$

which means $k \geqslant f(x)=\frac{2|x-1|}{x^{2}+6}$,

Therefore, $k \geqslant \max f(x), x \in \mathbf{R}$.
When $x \geqslant 1$, let $t=x-1, t \geqslant 0$, then
$$
\begin{array}{l}
f(x)=\frac{2 t}{(t+1)^{2}+6}=\frac{2 t}{\left(t^{2}+7\right)+2 t} \\
\leqslant \frac{2 t}{2 \sqrt{7} t+2 t}=\frac{\sqrt{7}-1}{6},
\end{array}
$$

with equality holding if and only if $t=\sqrt{7}$, i.e., $x=1+\sqrt{7}$.
When $x<1$, let $t=1-x, t>0$, then
$$
\begin{array}{l}
f(x)=\frac{2 t}{(1-t)^{2}+6}=\frac{2 t}{\left(t^{2}+7\right)-2 t} \\
\leqslant \frac{2 t}{2 \sqrt{7} t-2 t}=\frac{\sqrt{7}+1}{6},
\end{array}
$$

with equality holding if and only if $t=\sqrt{7}$, i.e., $x=1-\sqrt{7}$.
In summary, $\max _{x \in \mathbf{R}} f(x)=\frac{1+\sqrt{7}}{6}$, hence the range of $k$ is
$$
k \geqslant \frac{1+\sqrt{7}}{6} .
$$