# Uniform convergence of $f_{n}(x)=\frac{\log(1+nx^2)}{2n} \: \forall x\in[1,2]$

Given : $f_{n}(x)=\frac{\log(1+nx^2)}{2n} \: \forall x\in[1,2].\:$Prove it is uniformly convergent on the given interval. $\\$

I proved it is uniformly convergent using the following definition of preservance of differentiation $\colon \\$

$f_{n}(1)$ converges pointwise and $f_{n}^{\prime}(x)$ converges uniformly $\forall x \in [1,2] \Rightarrow f_{n}(x)$ converges uniformly $\forall x\in [1,2]$ and also preserves differentiation. $\\$

But if I try to solve using the log inequality, and then use the lim sup definition of uniform convergence , i don't get the answer (which means i can't use the inequality, but i am not able to visualize why), i.e. $\colon \\$

$$\log(t)\le (t-1) \quad \forall t>0 \\$$

$\therefore$ Assuming $1+nx^2=t$ and as $n\in \mathbb{N},\:x\in[1,2]\:\therefore1+nx^2>0 \quad$ and using the above inequality we get$\quad \colon \\$

$$\log( 1+nx^2)\le(1+nx^2-1)=nx^2\\$$

Also $\lim_{n\to \infty}f_{n}(x)=0 \:,$thus if $f_{n}(x)$ is uniformly convergent, it must converge to 0.Now using lim sup definition : $\\$

$$\lim \sup\{|f_{n}(x)-f(x)|:x\forall[1,2]\} \Rightarrow \lim \sup\left\{\left|\frac{\log(1+nx^2)}{2n}-0\right|:x\forall[1,2]\right\} \\$$

$$\left|\frac{\log(1+nx^2)}{2n}\right| \le \frac{nx^2}{2n}=\frac{x^2}{2}=2 \: \because x\in[1,2]\\$$

and $\lim_{n\to \infty} 2=2\ne 0$ and thus not uniformly convergent.

But I got it to be uniformly convergent using preservance in differentiation. I guess because $\log(1+nx^2)$ is a function in two variables x and n, we can't use the log inequality. I am sure its some basic intuition I am missing. can someone please clear this doubt. Many thanks.

• You only showed that $\limsup |f_n-f|\leq 2$, not equal to zero, so no problems there... In fact, you can show that it approaches zero as $n \to\infty$. – Yes May 7 '18 at 4:30
• got it.thanks for the help. – under_root May 7 '18 at 19:22

The function $\log$ is increasing, hence, for $x \in [1,2]$:
$\log(1+nx^2) \le \log(1+4n) \le \log(5n) = \log (5) + \log(n)$
Now use that $\frac{ \log (x)}{x} \to 0$ as $x \to \infty$.