Pointwise and uniform convergence of $f_n(x)=\sqrt[n]{nx^2+1}$

Question

Study the pointwise and uniform convergence of $f_n(x)=\sqrt[n]{nx^2+1}$ for $x\in\mathbb{R}$.

My attempt: let $x_0 \in \mathbb{R}$ fixed, it is $$\lim_{n\to\infty} \sqrt[n]{nx^2_0+1}=1$$ So $f_n$ converges pointwise to $f(x)=1$ for all $x\in\mathbb{R}$.

For the study of the uniform convergence, I have to evaluate $$\lim_{n \to \infty} \sup_{x \in \mathbb{R}}|\sqrt[n]{nx^2+1}-1|$$ The $n$-th root is an increasing function, so since $nx^2+1 \geq 1$ for all $x\in\mathbb{R}$ it is $\sqrt[n]{nx^2+1}-1 \geq 0$ and so $|\sqrt[n]{nx^2+1}-1|=(\sqrt[n]{nx^2+1}-1)$; moreover, it is $$\frac{\text{d}}{\text{d}x}(\sqrt[n]{nx^2+1}-1)=\frac{2x}{n}(nx^2+1)^{\frac{1-n}{n}} \geq 0 \iff x \geq 0$$ So the function $f_n-f$ is increasing function of $x$ and it is an unbounded function of $x$, since $$\lim_{x \to \infty} (f_n(x)-f(x)) = \lim_{x \to -\infty} (f_n(x)-f(x)) =\infty$$ So it is $$\sup_{x \in \mathbb{R}}(\sqrt[n]{nx^2+1}-1)=\infty$$ So $\lim_{n \to \infty} \sup_{x \in \mathbb{R}} |\sqrt[n]{nx^2+1}-1|=\infty$, this means that the $f_n$ doesn't converge uniformly to $f(x)=1$ in $\mathbb{R}$. However, given $a>0$ and considered the interval $[-a,a]$, by the same argument of monotonicity it is $$\sup_{x \in [-a,a]} |\sqrt[n]{nx^2+1}-1|=\sqrt[n]{na^2+1}-1 \to 0 \ \text{as} \ n \to \infty$$ So $f_n$ converges uniformly to $f(x)=1$ in every compact subset $[-a,a]$ of $\mathbb{R}$.

My questions are the following:

1. When I study the positivity of the derivative, I omit the term $(nx^2+1)^{\frac{1-n}{n}}$ because I see it as an exponential and so I conclude it is always positive. I have in general problems with the $n$-th power and $n$-th roots, is this correct?

2. When I study the uniform converge in $\mathbb{R}$ I have to take the limit of a supremum which is already $\infty$; this never happened in my previous study, so I'm unsure about this step. It has some sense to take the limit of something that is already $\infty$?

3. The fact that the convergence is uniform in every compact subset of $\mathbb{R}$ but it can't be extended in $\mathbb{R}$ is because when we introduce $a>0$ for defining the interval $[-a,a]$ we are fixing $a$ and so we can conclude with the estimation on the supremum while the elements of $\mathbb{R}$ are considered elements that vary without a fixed bound (unlike $a$) and this makes the difference for the uniform convergence? Or is there a deeper reason why the uniform convergence fails in $\mathbb{R}$ but holds in every compact subset of $\mathbb{R}$?

Thank you.

Answer 1

When you look at the limit for fixed $x_0$, why aren't you justifying the limit you write? I don't think it's dead obvious that limit is $1$.

After that, when you try to find the supremum of $|\sqrt[n]{nx^2 +1} - 1|$ for $x \in R$, what is your first impression about how big you can make this expression, if you're allowed to choose $x$ any way you like?

Answer 2

to deduce the non-convergence in $\mathbb{R}$, which seems to take $$ x_n = \sqrt {\frac {n^n-1}{n}}$$ (deducted from imposing $nx^2+1 = n^n $). In this case $$ f_n(x_n)-1 = n-1 $$ and $$ \sup (f_n(x)-1) \geq f_n(x_n)=n-1 $$ and the limit: $$ \lim_{n \to \infty} \sup (f_n(x)-1) \geq \lim_{n \to \infty} (n-1) = \infty \neq 0 $$ implies non-convergence in $\mathbb{R}$.