Check the pointwise and the uniform convergence of $(f_n)_n:D\rightarrow \mathbb{R}$

Question

Check the pointwise and the uniform convergence of $(f_n)_n:D\rightarrow \mathbb{R}$ with

1. $\displaystyle{f_n(x)=\frac{nx}{1+n^2x^2} \ \text{ for } x\in D=[-1,1]}$

2. $\displaystyle{f_n(x)=\sqrt{\frac{1}{n^2}+x^2} \ \text{ for } x\in D=\mathbb{R}}$

3. $\displaystyle{f_n(x)=\sqrt[n]{x} \ \text{ for } x\in D=(0,1)}$

$$$$

I have done the following:

1. If $x=0$ then we have that \begin{equation*}f(0)=\lim_{n\rightarrow +\infty}f_n(0)=\lim_{n\rightarrow +\infty}\frac{n\cdot 0}{1+n^2\cdot 0^2}=\lim_{n\rightarrow +\infty}\frac{0}{1}=0\end{equation*} If $x\neq 0$ then we have that \begin{align*}f(x)&=\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}\frac{nx}{1+n^2x^2}=\lim_{n\rightarrow +\infty}\frac{n^2\cdot \frac{x}{n}}{n^2\left (\frac{1}{n^2}+x^2\right )}\\ & =\lim_{n\rightarrow +\infty}\frac{ \frac{x}{n}}{\frac{1}{n^2}+x^2}=\frac{ 0}{0+x^2}=0\end{align*} So we have pointwise convergence on $[-1,1]$ to $f(x)=0$. The derivative is equal to $f_n'(x) =\frac{n-n^3x^2}{(1+n^2x^2)^2}$. It holds that $f_n$ is decreasing on $\left [-1, -\frac{1}{n} \right )$ and $\left (\frac{1}{n}, 1\right ]$ and increasing on $\left [-\frac{1}{n}, \frac{1}{n} \right ]$. So we have that $\max f_n=f_n\left(\frac1n\right)=\frac{1}{1+1}=\frac{1}{2}$. It holds that $\displaystyle{\lim_{n\to\infty}\max f_n=\frac{1}{2}\neq 0}$. So we don't have uniform convergence on $[-1,1]$ to $f(x)=0$.

2. We have that \begin{equation*}f(x)=\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}\sqrt{\frac{1}{n^2}+x^2}=\sqrt{x^2}=|x|\end{equation*} So we have pointwise convergence on $\mathbb{R}$ to $f(x)=|x|$.

3. We have that \begin{equation*}f(x)=\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}\sqrt[n]{x}=\lim_{n\rightarrow +\infty}x^{\frac{1}{n}}=1\end{equation*} So we have pointwise convergence on $(0,1)$ to $f(x)=1$.

Is everything correct so far? Could you give me a hint what we could do for the uniform convergence of the sequences of functions at 2 and 3 ?

Do we consider a function $g(x)=f_n(x)-f(x)$ and check if its maximum converges to $0$ ?

Answer 1

What you did is fine, except that, in the case of the third sequence, you forgot that $0$ dos not belong to $(0,1)$. On this set $(f_n)_{n\in\Bbb N}$ converges pointwise to $1$.

In the case of the second sequence, the convergence is uniform. Differentiating, you can check that $\sqrt{\frac1{n^2}+x^2}-x$ is decreasing on $[0,\infty)$. So, its maximum is attained at $0$, where it takes the value $\frac1n$. By a similar argument, the maximum of $\sqrt{\frac1{n^2}+x^2}+x$ on $(-\infty,0]$ is also $\frac1n$. So,$$\sup\left|\sqrt{\frac1{n^2}+x^2}-|x|\right|=\frac1n$$and therefore the convergence is uniform

The third sequence doesn't converge uniformly to $1$, because$$\lim_{n\to\infty}f_n\left(\frac1{n^n}\right)=\lim_{n\to\infty}\frac1n=0.$$