# Pick out true statements about the limit of $f_n(x)=\frac{1}{1+n^2x^2}$

For the sequence of functions $f_n(x)=\frac{1}{1+n^2x^2}$ for $n \in \mathbb{N}, x \in \mathbb{R}$ which of the following are true?

(A) $f_n$ converges point-wise to a continuous function on $[0,1]$

(B) $f_n$ converges uniformly on $[0,1]$

(C) $f_n$ converges uniformly on $[\frac{1}{2},1]$

(D) $\lim\limits_{n \to \infty} \int_0^1 f_n(x)dx=\int_0^1\lim\limits_{n \to \infty} f_n(x) dx$

So here is my take on this and I want to know if I am correct. Me and a friend are having quite a debate on this question and I can't see how he can disagree with the following logic.

Obviously A is false because the limit of $f_n$ on $[0,1]$ is $$F(x) =\begin{cases} 1 & if \ x=0\\ 0 & otherwise \end{cases}$$

Statement B is also false because if $f_n$ were to converge uniformly on $[0,1]$ then its limit would have been a continuous function which is obviously not the case.

C is true. This is because the uniform norm $||f_n - 0||$ converges to zero.

D is false because $$\lim\limits_{n \to \infty} \int_0^1 f_n(x)dx=\lim\limits_{n \to \infty} \int_0^1 \frac{1}{1+n^2x^2}dx=\frac{\pi}{2}$$ But $$\int_0^1\lim\limits_{n \to \infty} f_n(x) dx = 0$$

• You are right . But why do you think C) is true? – thedilated Jun 19 '17 at 5:43
• @thedilated Its simple application of Weierstrass M test. (added) – Castor Godinho Jun 19 '17 at 5:47
• Alternatively, you can simplify the argument by saying that the uniform norm of $|f_n - F|$ converges to 0. – thedilated Jun 19 '17 at 5:48
• @thedilated Yes, that's much better. Thanks! – Castor Godinho Jun 19 '17 at 5:49
• I think there's a mistake in D - $\int_0^1 f_n(x)\,dx = \frac{1}{n} \tan^{-1}(n) \to 0$ as $n\to\infty$. (Or, the sequence satisfies the conditions of the dominated convergence sequence since $|f_n(x)| \le 1$.) – Daniel Schepler Jun 19 '17 at 5:51

Let $\varepsilon>0$ be given. We observe that $\{\,f_n\}_{n=1}^\infty$ is a monotonically decreasing sequence of continuous functions $\,f_n: [0,1] \to \mathbb{R} \; (n=1,2,3,\ldots)$, and $$\int_{0}^{1}f_1(x) dx < \infty \, .$$ Thus we may find $t \in (0,1)$ so that \begin{align} \int_0^t f_1(x) dx \leq\frac{\varepsilon}{2} \, . \end{align}
By Dini's Theorem, we now have that $f_n \to 0$ uniformly on $[t, \, 1]$. Hence there is a positive integer $N$ so that \begin{align} |\,f_n(x)| < \frac{\varepsilon}{2(1-t)} \, \text{ whenever } n \geq N\text{ and } x \in [t, \, 1] \, . \end{align}
Therefore \begin{align}\left|\int_0^1 f_n(x)dx - \int_0^1 0 \,dx\right| &\leq \left|\int_t^1 f_n(x)dx\right| + \left|\int_0^t f_n(x)dx\right| \\& \leq \left|\int_t^1 f_n(x)dx\right| + \left|\int_0^t f_1(x)dx\right| \\& < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \end{align} whenever $n \geq N$.