If $\lim \int \phi f_n$ exists for every continuous $\phi$, does $\lim \int \phi \sqrt{1+f_n^2}$ exist?

Question

Question: Let $\{f_n \}_{n=1}^\infty$ be a sequence of non-negative real-valued continuous functions on $(0,1) \subset\mathbb R$ and such that for any continuous $\phi: (0,1) \rightarrow \mathbb R$ with compact support, the limit $$ \lim_{n\rightarrow \infty} \int \phi f_n $$ exists.

Then, is it true that for any $\phi: (0,1) \rightarrow \mathbb R$ with compact support, the limit $$ \lim_{n\rightarrow \infty} \int \phi \sqrt{1+(f_n)^2} $$ exists?

Attempt I tried triangle inequality, rewriting the integral, but just can't seem to get it. I think if the $f_n$ converge in measure that might be enough. It looks like it should be true but maybe it's not?

Answer 1

Nonlinear transformations tend to mess up weak convergence. Consider the sequence $$g_n(x) = 1+\operatorname{sign} \sin 2\pi nx$$ which is a fancy way of saying "$g_n$ jumps between $0$ and $2$ very often".

The continuity of $\phi$ implies $\int \phi g_n\to \int \phi$ because the contributions of adjacent $0$- and $2$- intervals average out to $1$.

On the other hand, $\sqrt{1+g_n^2}$ jumps between $1$ and $\sqrt{5}$, averaging out to the golden ratio $(1+\sqrt{5})/2$. So, $\int \phi \sqrt{1+g_n^2} \to \frac{1+\sqrt{5}}{2} \int \phi $.

This isn't a counterexample yet, since the limit still exists. For a counterexample, let $$ f_{2n} = g_n,\qquad f_{2n-1}\equiv 1 $$
Now $\int \phi f_n\to \int \phi$ but the sequence $ \int \phi \sqrt{1+f_n^2} $ does not have a limit: it alternates between $\frac{1+\sqrt{5}}{2} \int \phi $ and $\sqrt{2} \int \phi $.