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.
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$.