(Strong) Lower semicontinuity w.r.t. $L^\infty$ norm Let $B_1$ denote the unit ball in $\mathbb R^d$ and set $\mathcal K := \{\eta \in C_c^\infty(B_1): \eta \ge 0, \int \eta = 1,  \vert \nabla \eta \vert \le M \}$ where $M>0$ is a fixed real number (large enough). 
Consider the following functional 
$$\tag{1}
\mathcal K \ni \eta \mapsto \int_{B_1}\big\vert \langle x, \nabla \eta(x)\rangle \big\vert \, dx. 
$$
I am trying to understand its behaviour w.r.t. uniform topology on $\mathcal K$, i.e. the one induced by the $L^\infty$ norm.

Question. Is the functional $(1)$ continuous (or, at least, lower semicontinuous) on the compact set $\mathcal K$? 

I am puzzled... are there any general results I can apply? I was about to try to use directly the definition, i.e. take a sequence $\eta_n \to \eta$ uniformly; but do not know what to do now, I have no information a part the uniform bound on $\nabla \eta_n$... and I do not see how to apply Fatou's lemma. 
Thanks. 
EDIT: another try... Seems it is continuous w.r.t. $s-W^{1,1}$ topology... But I am not sure and I do not know how to use this piece of information.
 A: It is lower semicontinuous, but not continuous with respect to the uniform topology.
A good reference concerning lower semicontinuity in $W^{1,p}(\Omega)$ for functionals of the type
$$F(u)=\int_\Omega f(x,u(x),\nabla u(x))dx$$
is the book by Dacorogna, Direct Methods in the Calculus of Variations. In particular Corollary 3.22 apply to your case. As a general rule, in order to have weak-$W^{1,p}$ lower semicontinuity of these kind of functionals, the essential thing is that $f(x,s,\xi)$ must be convex in $\xi$. There are other assumptions on the semicontinuity and growth of $f$ etc., but they are usually verified for, say, $f$ continuous.
In your case the uniform convergence implies for any $p\geq 1$ the $L^p$ convergence which, together with the boundedness of the gradients in $L^p$,  is equivalent for $1<p<\infty$ to the weak-$W^{1,p}$ convergence (up to subsequences). Moreover $|\langle x,\xi\rangle|$ is convex in $\xi$, so you obtain the result. As noted in the comments, the lower semicontinuity result is still true in the weak*-$W^{1,\infty}$ topology. So we conclude if we can prove that up to subsequences $\eta_n\to \eta$ in this topology. But this is true by basically the same argument that proves the above result about weak convergence for $1<p<\infty$: by hypothesis $\eta_n\to \eta$ in $L^\infty$, and $\|\nabla \eta\|_\infty$ are bounded, so by Banach-Alaoglu they converge weakly* up to subsequences, and we are done.
Regarding the continuity, let us restrict to the case $d=1$, the general case being similar, and let us take functions in $W^{1,\infty}$ (the smooth case is obtained by approximation). Take for instance $\eta(x)=1-|x|$. The sequence $\eta_n$ is obtained with functions whose derivative in $(-1,0)$ oscillates between two values $a$ and $b$, and this happens on smaller and smaller scales as $n\to\infty$ (see picture which is zoomed in a point where $\eta'=1$; on $(0,1)$ the slopes are inverted).
$\hskip2in$ 
To have a global average slope of $1$, the condition is $b=2-a$. If both $a$ and $b$ are positive then the integral is $\frac12|a|+\frac12|2-a|=1$, and the functional converges to the right thing along the sequence. But if you take $a<0$ (as in the picture) the integral becomes $\frac12|a|+\frac12|2-a|=1+|a|$, and thus along this sequence the functional has a value strictly larger than the value in the limit function $\eta$. 
Instead if we take the strong $W^{1,1}$ topology you can check just by triangle inequalities that the functional is continuous.
