On uniform convergence of $f_n(x) = (2^n +|x|^n)^{1/n}$ for $x\in \mathbb{R}$

Question

Let $$f_n(x) = (2^n +|x|^n)^\frac{1}{n}, ~~~x\in \mathbb{R}$$

My question is: Does the sequence of ${f_n}$ converge uniformly on $\mathbb{R}$?

I was thinking about $M_n$ test but could not be able to find it out?

Any hint or solution would be appreciated.

Answer 1

Answer: From this Show that $\lim_{n\to\infty} \sqrt[n]{a^{n} + b^{n}} = \max(a, b)$ The pointwise convergence says, $$\color{blue}{\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty}(2^n +|x|^n)^\frac{1}{n} = \max(2,|x|):= f(x)}$$ and I have shown below that $$\color{red}{\sup_{x\in \Bbb R}|f_n(x)-f(x)| = 2(2^{1/n}-1)\to 0}$$

this prove that the uniform convergence.

Let us check the uniform convergence,

• For $|x|\ge 2$ we have $f(x)= |x|$ and let $$ g_n(t)= f_n(t) -f(t) = (2^n +t^n)^{\frac{1}{n}}-t ,~~~~~ t\ge 2$$$$\implies g'_n(t) = t^{n-1}(2^n +t^n)^{-\frac{n-1}{n}}-1 $$

the equation $g_n'(t) = 0 \Longleftrightarrow t^n = (2^n +t^n)$ has no solution in $(2,\infty)$.

hence $g_n$ therein has no critical point and since $\lim_{t\to \infty}g_n(t)= 0$. we have $$\color{red}{\sup_{0\le t\le 2}|g_n(t)| = \max(|g_n(\infty)|,|g_n(2)| )= 2(2^{1/n}-1)}$$

• Now, for $|x|\le 2 $ we have $f(x) = 2$ and we also let

$$h_n(t) = f_n(t)-f(x) = (2^n +|x|^n)^\frac{1}{n}-2,~~~0\le t<2$$ $$\implies h'_n(t) = t^{n-1}(2^n +t^n)^{-\frac{n-1}{n}}$$

Which only critical point is $t= 0$ then we have that (with $h_n(0)=0$) $$\color{red}{\sup_{0\le t\le 2}|h_n(t)| = \max(|h_n(0)|,|h_n(2)| )= 2(2^{1/n}-1)}$$

Answer 2

Lemma: For every $a\geqslant b\geqslant0$, $a\ne0$, and for every $n\geqslant2$, $$a\leqslant(a^n+b^n)^{1/n}\leqslant a+b^2/(na)$$

As a consequence, for every $x$ and every $n\geqslant2$, $$M(x)\leqslant(2^n+|x|^n)^{1/n}\leqslant M(x)+m(x)^2/(nM(x))$$ where $$m(x)=\min\{2,|x|\}\qquad M(x)=\max\{2,|x|\}$$ in particular,

$$M(x)\leqslant(2^n+|x|^n)^{1/n}\leqslant M(x)+(2/n)$$

which implies the desired uniform convergence.

The only part of the lemma whose proof is not direct is the inequality $$a^n+b^n\leqslant(a+b^2/(na))^n$$ which is equivalent to the fact that, for every $u$ in $[0,1]$, $$1+u^n\leqslant(1+u^2/n)^n$$ but this is direct since, for every $n\geqslant2$, $$(1+u^2/n)^n\geqslant1+u^2\geqslant1+u^n$$

Answer 3

Note first that the function defined by $\frac{\exp(u)-1}{u}$ for $u\in ]0,1]$ and $1$ at $0$ is continuous on $[0,1]$, hence bounded, say by $c>0$ and that for $x\geq 0$ we have $\log(1+x)\leq x$.

Put on $[0,1]$, $g_n(v)=(1+v^n)^{1/n}=\exp(\frac{\log (1+v^n)}{n})$. Applying the above inequalities, we find that for $v\in [0,1]$ $$|g_n(v)-1|\leq c\frac{v^n}{n}$$ Now, for $|x|\leq 2$, we have $f_n(x)=2g_n(\frac{|x|}{2})$, hence $$|f_n(x)-2|\leq 2 c\frac{|x|^n}{n2^n}\leq \frac{2c}{n}$$

For $|x|\geq 2$, $f_n(x)=|x|g_n(\frac{2}{|x|})$

$$|f_n(x)-|x||\leq |x| c\frac{2^n}{n|x|^n}\leq 2c\frac{2^{n-1}}{n|x|^{n-1}}\leq\frac{2c}{n}$$

Hence $$|f_n(x)-\max\{2,|x|\}|\leq \frac{2c}{n}$$ for all $x\in \mathbb{R}$, and the convergence is uniform.