If $f(x,y)$ is continuous in $\mathbb{R}^2$, and it is also bounded. Define $\phi(y):=\|f^y(x)\|_{L_{\infty}}$ where $f^y(x)$ is the slice of the function $f$. I want to prove or disprove that $\phi(y)$ is continuous with respect to $y$.
My thoughts: For $y_n\to y$ we need $|\phi(y_n)-\phi(y)|\to0$ to show $\phi$ is continuous. We know $|\phi(y_n)-\phi(y)|=|\|f^{y_n}(x)\|_{L_{\infty}}-\|f^y(x)\|_{L_{\infty}}|\le\|f^{y_n}(x)-f^{y}(x)\|_{L_{\infty}}$. Now, we know $f^{y_n}(x)\to f^{y}(x)$ pointwisely for any $x$. If we can show that this implies $\|f^{y_n}(x)-f^{y}(x)\|_{L_{\infty}}\to0$ then we will prove the claim. It is clear that, in general, $g_n(x)\to g(x)$ pointwisely doesn't imply $\|g_n(x)-g(x)\|_{L_{\infty}}\to0$. However, here we are not dealing with a general case, and I think I should use some other properties of slice of a continuous bounded function to conclude the proof.
There exists a counterexample, which I construct by means of general topology. Let $A=\{(x,y)\in\Bbb R^2: y=0\}$ and $B=\{(x,y)\in\Bbb R^2: |xy|\ge 1\}$. Then $A$ and $B$ are disjoint closed subsets of $\Bbb R^2$. Since the space $\Bbb R^2$ is metric, it is normal, so there exists a continuous function $f:\Bbb R^2\to [0,1]$ such that $f(A)=\{0\}$ and $f(B)=\{1\}$. Then $\phi(y)=0$, if $y=0$ and $\phi(y)=1$, otherwise.