Passing Weak Convergence Through a Function?

Question

Let $f_k,f : \mathbb{R}^d\to \mathbb{R}$. Assume $f_{k}$ converges weakly to $f$ in $L^p(\mathbb{R}^d)$. Let $p:\mathbb{R}\to \mathbb{R}$ be some function.

What can say about the weak convergence of $p(f_k)$ (what are some general assumptions on $p$ that one would expect to need)? The reason I ask is I have two terms in an equation in which I want to evaluate as $k\to\infty$, these are

$$ \int_{\mathbb{R}^d} \varphi_1(x)f_k(x)+ \varphi_2(x)p(f_k(x))~dx $$

where $\varphi_i\in C^\infty_c(\mathbb{R})$, infinitely differentiable compact support, i.e are test functions. I don't mind if I can evaluate this limit $k\to\infty$ up to a subsequence.

Answer 1

It seemed a good idea to me to assume $p$ as Lipschitz continuous for some nice resutlts, however in the comments I was shown that it doesn't really get you anywhere by itself, so I have edited the answer cancelling the wrong fact I had wrote, but still what follow works.

Also with the "famous" theorem (1.1 in here):

Theorem 1.1. $(1 < p < +\infty)$ Let $\{f_n\}$ be a bounded sequence in $\textit{L}^p(U)$. Then there exists a subsequence $\{f_{n_k}\}$ which converges weakly to some $f\in \textit{L}^p(U)$.

With this result you could also argue that, for weak convergence of at least a subsequence of $\{ p(f_k) \}$, you only need the boundedness of $||p(f_k)||_p$.

Answer 2

This is not possible unless $p$ is affine.

Indeed, if $p$ is not affine, then there are $a,b \in \mathbb R$ and $\lambda \in (0,1)$ such that $$ p( \lambda \, a + (1-\lambda) \, b) \ne \lambda \, p(a) + (1-\lambda) \, p(b).$$

Let us denote the support of $\varphi_2$ by $C$ and the Lebesgue measure by $\mu$. For simplicity, I will assume that $\varphi_2 = 1$ on $C$, otherwise the argument still works but is a little bit more technical.

Now, we can construct a sequence $f_k$, such that each $f_k$ only takes the values $a$ and $b$ on $C$ with $$ \mu(\{ x \in C \;|\; f_k(x) = a \}) = \lambda \, \mu(C)$$ and such that $f_k \rightharpoonup f$, where $f$ equals $\lambda \, a + (1-\lambda)\,b$ on the support of $\varphi_2$. (This is possible by some standard construction of oscillating sequences).

Then $$\int \varphi_2 \, p(f_k) \mathrm d\mu = (\lambda \, p(a) + (1-\lambda) \, p(b)) \, \mu(C),$$ but $$\int \varphi_2 \, p(f) \mathrm d\mu = p(\lambda \, a + (1-\lambda) \, b) \, \mu(C).$$