Uniform convergence of $f_n(x)=\frac{1}{1+nx}$ on $(0,1)$

Question

Consider the sequence of functions $f_n(x)=\frac{1}{1+nx}$ for $x\in (0,1)$. Then

1. $f_n (x) → 0$ pointwise but not uniformly on $(0,1)$.
2. $f_n (x) → 0$ uniformly on $(0,1)$.
3. $\int_{0}^{1} f_n (x) dx → 0$ as $n → ∞$.
4. $f'_n(1/n)\to 0$ as $n\to \infty$.

I have shown that $3$ and $4$ are false. As $\lim_{n\to \infty} \int_{0}^{1}f_n(x)dx=1$ and $f'_n(1/n)=-\frac{n}{4}$. But I am confused about the convergence part. I know $f_n(x)$ is not uniformly convergent on $[0,1]$, so can I just remove the end points and claim that it is also not uniformly convergent on $(0,1)$ as well. It does not feels rigorous. How can I do this. Please help me out.

Answer 1

Let $\epsilon=\frac{1}{2}$, $n_k = k+1$, and $x_k=\frac{1}{k+1}$, then \begin{align} |f_{n_k}(x_k)-f(x_k)| &= \left|\frac{1}{1+(k+1)\cdot\frac{1}{k+1}}\right|\\ &=\frac{1}{2}\\ &\ge \epsilon, \end{align} so $(f_n)$ does not uniformly converge.

Answer 2

Yes, $1/(1+nx) \to 0$ pointwise on $(0,1).$ If the convergence were uniform, then we would have $ \sup_{x\in (0,1)} 1/(1+nx) \to 0.$ But these suprema are $1$ for all $n.$ Thus the convergence is not uniform. You've made a mistake with the integrals: $\int_0^1 1/(1+nx)\, dx = [\ln (1+n)]/n \to 0.$

Answer 3

You are mistaken about #3. It is true. Fix any $a\in (0,1).$ For $n>0$ we have $$0<\int_0^1 f_n(x)\;dx=\int_0^a f_n(x)\;dx+\int_a^1f_n(x)\;dx<$$ $$<\int_0^a1\;dx+\int_a^1(1/1+na)\;dx=a+(1-a)/(1+na).$$ So for all $a\in (0,1)$ we have $0\leq \lim_{m\to \infty}\sup_{n>m}\int_0^1f_n(x)\;dx\leq a.$