# The sequences $\{f_n\},\{f_n'\}$ both are uniformly convergent on $[0,1]$.

Let $f_n(x) = \frac{\ln(1 + n^2x^2)}{n^2}, x \in [0,1]$. Then the sequences $\{f_n\},\{f_n'\}$ both are uniformly convergent on $[0,1]$.

Here, $f_n'(x) = \frac{2x}{1 + n^2x^2}$. Both limits of $f_n,f_n'$ when taken as $n \to \infty$ is $0$. I have verifyed that the $f_n'$ is a monotone decreasing function on $[0,1]$ and hence by Dini's Theorem we have $f_n'$ uniformly convergent on $[0,1]$.

But I have tried both $M_n$ test and Dini's Theorem on $f_n$ but unable to prove that $f_n$ is uniformly convergent on $[0,1]$. Please Help!

## 3 Answers

Since $$|f_n(x)|=\frac{\ln(1 + n^2x^2)}{n^2}<\frac{\ln(2n^2)}{n^2}=\dfrac{\ln{2}+2\ln{n}}{n^2}\to0,\hspace{3 mm} \forall x\in[0,1]$$ So by Weierstrass-M test, $\{f_n\}$ is uniformly convergent on $[0,1]$.

Also $$|f_n'(x)|=\dfrac{2x}{(1 + n^2x^2)}\leqslant |f_n'(\dfrac1{n})|=\dfrac{2}{n}\to0,\hspace{3 mm} \forall x\in[0,1]$$ for $|f_n{''}(\dfrac1{n})|=\dfrac{1-n^2x^2}{(1 + n^2x^2)^2}\bigg|_{1/n}=0$. So by Weierstrass-M test, $\{f'_n\}$ is uniformly convergent on $[0,1]$.

Consider the fact that $f_n(x) = \int_{0}^{x} f'_n(t) dt$. Thus, $|f_n(x)| \leq \int_0^x |f'_n|(t) dt$. Now apply what you know about uniform limits and integrals to get what you want.

Note that $$\ln (1+n^2x^2)\le \ln (1+n^2)\le \ln (n^2+n^2)= \ln (2n^2) = \ln 2 +2\ln n.$$

Thus for each $n$ and all $x\in [0,1],$

$$|f_n(x)| \le \frac{\ln 2 +2\ln n}{n^2}.$$

The limit of the last sequence is $0$ and that tells you $f_n \to 0$ uniformly on $[0,1].$

Another approach to showing $f_n' \to 0$ uniformly is to maximize

$$|f_n'(x)| = f_n'(x)=\frac{2x}{1+n^2x^2}$$

over $[0,1]$ by taking the derivative of $f_n'$ and setting it equal to $0.$ I get $1/n$ for this maximum, showing $f_n' \to 0$ uniformly.