Continuity of $\sup_{x\in\Omega}\varphi(x,\cdot)$ Let $\Omega\subset\mathbb{R}^n$ be open,bounded and (I don't know if this matter) of class $C^{1+\alpha}$. Let $\varphi:\bar\Omega\times\partial\Omega\to\mathbb{R}$ such that $\varphi(x,\cdot)$ is continuous on $\partial\Omega$ for all $x\in\bar\Omega$. 
Let this quantities be finite:
\begin{equation}
  \sup_{x\in\bar\Omega}||\varphi(x,\cdot)||_{C^0(\partial\Omega)}
\end{equation}
\begin{equation}
 \sup_{x_1\neq x_2, x_1,x_2\in\bar\Omega}
  \frac{||\varphi(x_1,\cdot)-\varphi(x_2,\cdot)||_{C^0(\partial\Omega)}}{|x_1-x_2|^\alpha}.
\end{equation}
It is true that $\sup_{x\in\Omega}\varphi(x,\cdot)$ is continuous on $\partial\Omega$?
I suppose that some hypothesis, like the second quantity requested to be finite, don't matter for my question.
 A: The finiteness of
$$M := \sup_{x_1\neq x_2, x_1,x_2\in\bar\Omega}
  \frac{\lVert\varphi(x_1,\cdot)-\varphi(x_2,\cdot)\rVert_{C^0(\partial\Omega)}}{\lvert x_1-x_2\rvert^\alpha},\tag{1}$$
assuming $\alpha > 0$, shows that $\varphi$ is actually continuous on $\overline{\Omega}\times \partial \Omega$:
\begin{align}
\lvert\varphi(x_1,y_1) - \varphi(x_2,y_2)\rvert &= \lvert \varphi(x_1,y_1) - \varphi(x_1,y_2) + \varphi(x_1,y_2) - \varphi(x_2,y_2)\rvert\\
&\leqslant \lvert \varphi(x_1,y_1)-\varphi(x_1,y_2)\rvert + \lVert \varphi(x_1,\cdot) - \varphi(x_2,\cdot)\rVert_{C^0(\partial\Omega)}\\
&\leqslant  \lvert \varphi(x_1,y_1)-\varphi(x_1,y_2)\rvert + M\cdot \lvert x_1-x_2\rvert^\alpha.
\end{align}
Then, given $\varepsilon > 0$, the continuity of $\varphi(x_1,\cdot)$ says that there is a $\delta > 0$ such that $\lvert y_2 - y_1\rvert \leqslant \delta$ implies $\lvert \varphi(x_1,y_1)-\varphi(x_1,y_2)\rvert \leqslant \frac{\varepsilon}{2}$, and thus we have
$$\lvert \varphi(x_1,y_1) - \varphi(x_2,y_2)\rvert \leqslant \varepsilon$$
for $\lvert y_2 - y_1\rvert \leqslant \delta$ and $\lvert x_2 - x_1\rvert \leqslant \left(\frac{\varepsilon}{2M}\right)^{1/\alpha}$, so $\varphi$ is continuous at $(x_1,y_1)$.
Now we can use the fact that if $C,K$ are compact$^{(1)}$ spaces, and $f\colon C\times K \to \mathbb{R}$ is continuous, then
$$g\colon K\to \mathbb{R},\quad g(y) = \sup_{x\in C} f(x,y)$$
is also continuous.
Since any supremum of continuous functions [lower semicontinuous suffices] is lower semicontinuous, we only need to check upper semicontinuity of $g$. Now let $a\in\mathbb{R}$. We need to see that $W = g^{-1}((-\infty,a))$ is open. If $W$ is empty, we're done, otherwise pick an $y_0 \in W$. Then
$$U = f^{-1}((-\infty,a))$$
is an open set containing the compact set $C\times \{y_0\}$, and hence it contains an open set of the form $C\times V$, where $V$ is an open neighbourhood of $y_0$. But then we have $g(y) < a$ for all $y\in V$, so $V\subset W$, hence $W$ is a neighbourhood of $y_0$. Since $y_0\in W$ was arbitrary, it follows that $W$ is open, and since $a$ was arbitrary, $g$ is upper semicontinuous.
If we didn't have a condition like $(1)$ that ensures continuity of $\varphi$, we would basically have only a (uniformly bounded) family of continuous functions on $\partial\Omega$, and then we could only deduce lower semicontinuity of
$$\sigma := \sup_{x\in\overline{\Omega}} \varphi(x,\cdot),$$
so condition $(1)$ is not superfluous. Of course it could be replaced by other conditions ensuring the continuity of $\varphi$ on $\overline{\Omega}\times\partial\Omega$, or the continuity of $\sigma$ without the continuity of $\varphi$.
The regularity of $\partial \Omega$ on the other hand is irrelevant, except that it possibly was used to establish condition $(1)$.

$^{(1)}$ That means quasicompact and Hausdorff.
