# Uniform convergence of proportion of positive real numbers at which uniformly convergent sequence of functions is positive

For a continuous function $$g: \mathbb R \rightarrow \mathbb R$$, we define the function $$p_g: \mathbb R^+ \rightarrow \mathbb R^+$$ by $$p_g(T) := \frac 1T\lambda(g^{-1}((0, \infty)) \cap [0, T])$$ where $$\lambda$$ denotes the Lebesgue measure.

Let $$f_n: \mathbb R \rightarrow \mathbb R$$ be a sequence of continuous functions which uniformly converges to a continuous function $$f: \mathbb R \rightarrow \mathbb R$$. What I am interested in finding out is whether the sequence of functions $$\{p_{f_n}\}_{n=1}^\infty$$ converges uniformly to $$p_f$$ on say, some $$[A, \infty)$$ for some fixed positive real number $$A$$. If necessary, we may assume the $$f_n$$'s and $$f$$ to be smooth.

Writing $$\lambda(g^{-1}((0, \infty)) \cap [0, T]) = \int_0^T \mathbbm{1}(g(t)>0) \, \text dt,$$ the Dominated Convergence Theorem shows that $$p_{f_n}(T)$$ converges to $$p_f(T)$$ pointwise for each $$T>0$$. However, I am having some trouble getting the uniform convergence I want. It may be useful to note that $$p_g(T)$$ is just the probability of $$x \in [0, T]$$ of having $$p_g(x)>0$$.

I feel like this should be fairly straightforward and should hold in greater generality (say, without any assumptions besides the uniform convergence $$f_n \rightarrow f$$ and, perhaps, Lebesgue measurability of the functions $$f_n$$ and $$f$$). I fear I may be missing some really simple idea; I haven't found any result of this form anywhere either, even though it looks fairly standard. Hence, I would be really grateful for any suggestions or references, even if one needs to assume that $$f_n$$'s and $$f$$ are smooth.

P.S.: In addition, what are the minimal conditions on the $$f_n$$ and $$f$$ for this to hold?

• I don't really have a full answer, but if the sequence converges point wise (which I am not totally sure it does, maybe your proof regarding that would be enlightening) and you manage to show that the sequence is monotone, then by Dini's Theorem, the sequence convergences compactly. Jan 16, 2022 at 13:59
• Further food for thought on any compact set where $f$ is either positive or negative, the sequence is either positive or negative on that exact same set in finitely many steps. So it seems the issue seems to revolve around the zeros of $f$ and their neighborhoods, however I haven't managed to wrap my head around it yet. Jan 16, 2022 at 14:09

I have some results here: First of all, I am going to write $$g^{-1}((0,\infty))$$ as $$\{g>0\}$$.
Second of all, I don't think mere convergence of $$f_n$$ to $$f$$ is enough to use Dominated Convergence to show that $$p_{f_n}\to p_f$$. As you correctly point out you can rewrite $$p_g(T)=\frac{1}{T}\int \unicode{x1D7D9}_{[0,T]\cap\{g>0\}}d\lambda$$ and the sequence is obiviously dominated by $$\unicode{x1D7D9}_{[0,T]}$$ which is obviously integrable, but in order to use Dominated Convergence on this form you would have to show that $$\unicode{x1D7D9}_{[0,T]\cap\{f_n>0\}}\to\unicode{x1D7D9}_{[0,T]\cap\{f>0\}}$$ pointwise first.
Let $$x\in[0,T]$$. Then $$|\unicode{x1D7D9}_{[0,T]\cap\{f_n>0\}}(x)-\unicode{x1D7D9}_{[0,T]\cap\{f>0\}}(x)|$$ is either $$0$$ or $$1$$. If $$f(x)>0$$ and $$f_n$$ converges pointwise to $$f$$ then there exists an $$n$$ s.t. $$f_n(x)>0$$ and the absolute difference is therefore $$0$$. Same deal if $$f(x)<0$$. However, if $$f(x)=0$$ then the absolute difference might always be $$1$$ if the convergence is from above, right? $$f_n(x)$$ could always be positive, but converge to $$0$$.
So, all you need is, if $$f$$ is measurable and $$f_n\to f$$ pointwise such that the zeros of $$f$$ are approached from below then $$p_{f_n}\to p_f$$ pointwise.
Now, notice that for any measurable function $$g$$, $$p_g$$ is continuous. If you look at the function $$L(t)=\lambda(A\cap[0,t])$$ for some set of real numbers $$A$$, you get for any $$x>y>0$$ $$|L(x)-L(y)|=|\lambda(A\cap[y,x])|\leq x-y$$ which means it is even Lipschitz-continuous. Using $$A=\{g>0\}$$ and multiplying by the multiplicative inverse function, gets you $$p_g$$ which is also continous since those functions are closed under multiplication.
Meaning, if $$f$$ is measurable and $$f_n$$ is a monotonically increasing sequence which converges pointwise to $$f$$, we get by the result from before that $$p_{f_n}\to p_f$$ pointwise. Furthermore, the sequence $$p_{f_n}$$ is also monotonically increasing and $$p_{f_n}$$ and $$p_f$$ are continuous and therefore by Dini's Theorem $$p_{f_n}\to p_f$$ compactly.