$$\text{ Let } f_n(x)=n^px(1-x^2)^n, x \in [0,1], \text{ where } p \in \mathbb{R} \text{ a parameter } \in \mathbb{R}. $$
$$\text{ Prove that } \forall p, (f_n) \text{ converges pointwise to an f in [0,1]. }$$ $$\text{ For which values of p,is the convergence uniform? For which values of p do we have:} $$ $$\int_0^1 f_n(x)dx \to \int_0^1 f(x)dx \text{ ?}$$
In order to have uniform convergence,it must be like that: $$\sup_{x \in [0,1]} |f_n(x)-f(x)| \to 0 \Rightarrow \sup_{x \in [0,1]} |n^px(1-x^2)^n| \to 0 \Rightarrow \frac{n^p}{\sqrt{1+2n}} \sqrt{(1-\frac{1}{1+2n})^{2n+1}(1+\frac{1}{2n})} \to 0\Rightarrow \frac{1}{n^{\frac{1}{2}-p}\sqrt{2+\frac{1}{n}}} \to 0 \Rightarrow p<\frac{1}{2}$$
In order to have:$\int_0^1 f_n(x)dx \to \int_0^1 f(x)dx$
it must be:
$$\int_0^1 n^px(1-x^2)^ndx=\int_0^1 0dx=0 \Rightarrow \frac{-n^p}{2} \int_0^1 x^ndx=0 \dots \Rightarrow p<1$$
But why are the values that we found for $p$ different at the above cases?
Shouldn't we found the same,according to the theorem:
$$\text{Let } f_n \to f \text{ uniformly in } [a,b]. \text{ If each } f_n \text{ is integrable in } [a,b], \text{ then } f \text{ is also integrable in [a,b] and } \int_a^b f_n(x)dx \to \int_a^b f(x)dx $$
Or am I wrong?
