Consider the case $\kappa = 0$, $1 < s < \lambda \leqslant 1$, the general one can be easily deduced from this one. Firstly, because $h$ is bounded in $\overline{\Omega}$ we have trivially: $\|
hu \|_p \leqslant \| h \|_{\infty} \| u \|_p$. Secondly, for the Gagliardo
seminorm
$$ [ hu ]_{s, p}^p := \underset{\Omega \times
\Omega}{\int \int} \frac{| h (x) u (x) - h (y) u (y) |^p}{| x - y |^{n +
sp}} \mathrm{d}x \mathrm{d}y, $$
we use the convexity of $x \mapsto x^p$ twice (that is, we use $| a + b |^p
\leqslant 2^{p - 1} (| a |^p + | b |^p)$ and obtain
\begin{eqnarray*}
\hspace{1em} {| h (x) u (x) - h (y) u (y) |^p} & \leqslant & C \left( | h
(x) u (x) - h (x) u (y) |^p + | h (x) u (y) - h (x) u (x) |^p \\
\hspace{2em} + | h (x) u (x) - h (y) u (x) |^p + | h (y) u (x) - h (y) u
(y) |^p \right)\\
& = & C \left( | h (x) |^p | u (x) - u (y) |^p + | h (x) |^p | u (y) - u
(x) |^p \\
\hspace{2em} + | h (x) - h (y) |^p | u (x) |^p + | h (y) |^p | u (x) - u
(y) |^p \right) .
\end{eqnarray*}
We plug this into the integral but for the third summand we use first that
$$ | h (x) - h (y) |^p \leqslant C | x - y |^{\lambda p}, $$
then:
\begin{eqnarray*}
[ hu ]_{s, p}^p & \leqslant & C \left( \| h \|_{\infty}^p
3 \underset{\Omega \times \Omega}{\int \int} \frac{| u (x) - u (y) |^p}{| x
- y |^{n + sp}} \mathrm{d}x \mathrm{d}y + \underset{\Omega \times \Omega}{\int
\int} \frac{C | x - y |^{\lambda p} | u (x) |^p}{| x - y |^{n + sp}} \mathrm{d}x
\mathrm{d}y \right)\\
& \leqslant & C \left( [ u ]_{s, p}^p + \underset{\Omega
\times \Omega}{\int \int} \frac{ | u (x) |^p}{| x - y |^{n + (s - \lambda)
p}} \mathrm{d}x \mathrm{d}y \right) .
\end{eqnarray*}
Notice that the exponent in the denominator of the integrand is $n - \delta <
n$ for some $\delta > 0$ so the function $1 / | z |^{n - \delta}$ is
integrable in any bounded domain. We use Fubini-Tonelli and change the
variable in the $\mathrm{d}y$ integral with $z = x - y$ to obtain:
\begin{eqnarray*}
[ hu ]_{s, p}^p & \leqslant & C \left( [ u
]_{s, p}^p + \int_{\Omega} | u (x) |^p \int_{x - \Omega}
\frac{1}{| z |^{n - \delta}} \mathrm{d}z \mathrm{d}x \right)\\
& \leqslant & C \left( [ u ]_{s, p}^p + \int_{\Omega} | u
(x) |^p \int_{\Omega + \Omega} \frac{1}{| z |^{n - \delta}} \mathrm{d}z \mathrm{d}
x \right)\\
& \leqslant & C ([ u ]_{s, p}^p + \| u \|_p^p) .
\end{eqnarray*}
Therefore
$$ \| hu \|_{s, p} \leqslant C \| u \|_{s, p}, $$
and keeping track of the constants above we see that indeed $C = C (h, p,
\Omega)$.
The factor $3$ is not relevant, so let us consider $f(x) = x^{1/3}$ instead. ($f$ is defined on $\Bbb R$ as the inverse function of $y \mapsto y^3$).
$f$ is indeed Hölder continuous with exponent $\alpha = \frac 13$, as you suspected. In order to prove an estimate
$$
|f(x) - f(y) | \le C |x-y|^{1/3}
$$
for all $x, y \in \Bbb R$ we distinguish two cases:
Case 1: $x, y$ have the same sign. Without loss of generality we can assume that $x, y \ge 0$. In this case,
$$
|f(x) - f(y) | \le |x-y|^{1/3}
$$
as shown in proving that $f(x) = x^s$ is holder continuous with holder exponent s.
Case 2: $x, y$ have opposite sign. Without loss of generality $x < 0 < y$. Then
$$
|f(x) - f(y) | \le C |x-y|^{1/3} \\
\iff |x|^{1/3} + y^{1/3} \le C (|x| + y)^{1/3} \\
\iff \left( \frac{|x|}{|x| + y}\right)^{1/3} + \left( \frac{y}{|x| + y}\right)^{1/3} \le C
$$
Now both fractions on the left-hand side are less than one, so that the estimate is satisfied with $C=2$:
$$
|f(x) - f(y) | \le 2 |x-y|^{1/3}
$$
This completes the proof of the Hölder continuity.
One can also determine the best possible constant $C$ by computing the maximum of
$$
u^{1/3} + (1-u)^{1/3} \text{ for } 0 < u < 1 \, ,
$$
which turns out to be $C = 2^{2/3}$.
For the (non) Lipschitz continuity, consider
$$
\frac{|f(x)-f(0)|}{|x-0|}
$$
for $x \ne 0$, $x \to 0$.
Best Answer
Your condition is $q = \alpha p$, so that $$\int (|f|^\alpha)^p = \int |f|^{\alpha p} = \int |f|^q.$$ Thus $$\|f^\alpha\|_p = \left( \int (|f|^\alpha)^p \right)^{1/p} = \left( \int |f|^q \right)^{1/p} = \left( \int |f|^q \right)^{\alpha/q} = \|f\|^\alpha_q.$$