$f_n$ defined on $[-1,1]$ by $f_n(x) = |x|^{1+\frac1n}, \ x \in [-1,1]$ converges uniformly .

Question

To show that the sequence of function $f_n$ defined on $[-1,1]$ by $f_n(x) = |x|^{1+\frac1n}, \ x \in [-1,1]$ converges uniformly to a function $f(x) = |x|, \ x \in [-1,1]$ and the sequence of function $f_n^{'}$ converges pointwise ( but not uniformly) on $[-1,1]$ to a function $g(x) = sgn x , x\in [-1,1]$.

$\lim_{n\to \infty} |x|^{1+\frac1n} = |x|$. Thus $f_n$ converges to a function $f(x) = |x|, \ x \in [-1,1]$. And also we see that $f_n^{'}(x) = (1+\frac1n)x^{\frac1n} , x> 0\ ; = -(1+\frac1n)(-x)^{\frac1n}, x<0\ ; = 0, x = 0$.

What would you suggest me to do here? Thanks.

Answer 1

HINT: PART 1:

You may use Dini's Theorem to show that $f_n$ converges uniformly to $f$.

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PART 2:

For the second, let's look at

$$\begin{align} \left|f_n'(x)-f'(x)\right|&=\left|\left(1+\frac{1}{n}\right)|x|^{1/n}\text{sgn}(x)-\text{sgn}(x)\right|\\\\ &=|\text{sgn}(x)|\left|\left(1+\frac{1}{n}\right)|x|^{1/n}-1\right|\\\\ \end{align}$$

Now, let's choose $x=\frac{1}{2^{n}}$. Then, we have

$$\begin{align} \left|f_n'(x)-f'(x)\right|&=|\text{sgn}(x)|\left|\left(1+\frac{1}{n}\right)|x|^{1/n}-1\right|\\\\ &=\left|\frac12 \left(1+\frac{1}{n}\right)-1\right|\\\\ &\ge \frac14 \end{align}$$

for $n\ge2$. Thus, there exists an $\epsilon>0$ (here we may take $\epsilon =1/4$), such that for any $N$ there exists $n>N$ and a point $x$ (here we take $x=1/2^n$), we have

$$\begin{align} \left|f_n'(x)-f'(x)\right|\ge \epsilon \end{align}$$

This is the negation of uniform convergence.

Answer 2

The function $\phi_n:[0,1]\to [0,1], \, \phi_n(t)=t-t^{1+\frac1n}$ has a maximum at the point $a_n\in [0,1]$ satisfying $$ \phi_n'(a_n)=1-\left(1+\frac1n\right)a_n^{\frac1n}, $$ i.e. $$ a_n=\left(\frac{n}{n+1}\right)^n, $$ and $$ \phi_n(a_n)=a_n(1-a_n^{\frac1n})=\frac{a_n}{n+1} $$ For every $x\in [-1,1]$ we have $|f(x)-f_n(x)|=\phi_n(|x|)$, and therefore $$ \max_{x\in [-1,1]}|f(x)-f_n(x)|=\phi_n(a_n)=\frac{a_n}{n+1}. $$ Since $$ \lim_{n\to\infty}\frac{a_n}{n+1}=\lim_{n\to\infty}\frac{1}{n+1}\cdot\left(1+\frac1n\right)^{-n}=0\cdot e^{-1}=0, $$ it follows that $f_n\to f$ uniformly on $[-1,1]$.