For $f\colon [-1,1]\to \mathbb{R}$ consider $$ \int_{-1}^1 (f(x) - f(-1))^2 \frac{1}{1+x}\, \mathrm{d}x. $$
How nice (how continuous, how regular, how ...) does $f$ have to be for this integral to be finite? Put another way, what is a broad class of functions for which this integral is finite?
As a follow up, does there exist a constant $C>0$ such that for all $f$ such that the integral is finite\ we have
$$\int_{-1}^1 (f(x)-f(-1))^2 \frac{1}{1+x}\, \mathrm{d} x \le C( |f(-1)| + \left\| f \right\|_{L^2})?$$
Thanks
First question: Here's the sufficient condition that @Dr.MV and I were discussing in the comments: Suppose $p>0,-1<b\le 1.$ Suppose $f$ satisfies
i) $|f(x)-f(-1)|\le C|x+1|^p$ in $[-1,b];$
ii) $f\in L^2[-1,1].$
Then the given integral is finite. Proof: i) shows the full integrand is $\le C^2(x+1)^{2p-1}$ on $[-1,b],$ which gives integrability over $[-1,b].$ ii) shows the rest of the integral is finite. (Note $f\in L^2[-1,1]$ implies $f\in L^1[-1,1].$ So when you expand the square to get $f(x)^2 -2f(x)f(-1) + f(-1)^2$ upstairs, the full integrand is integrable over $[-1,b].$)
This shows for example that any $f\in L^2[-1,1]$ that is differentiable at $-1$ yields a finite integral.
Follow-up question: The answer is no. Consider $f_n(x) =(1+x)^{1/n}$ to see this.