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The sequence $f_n \subseteq C[-1,1]$ given by $f_n(x) = \sqrt{\frac{1}{n^2}+x^2}$. Verify that each $f_n$ is differentiable on $(-1,1)$, and that $(f_n)$ converges in $C[-1,1]$ to the absolute value function. $C[-1,1]$ means the metric space with metric $d(f,g) = sup\{|f(x)-g(x)|: x \in [-1,1]\}$.

For verifying $f_n$ is differentiable on $(-1,1)$, is that mean for every $n \in N$ and $x \in (-1,1)$, $f_n'(x)$ exists? I have no idea how to show $f_n$ converges to the absolute value function in the $C[-1,1]$

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  • $\begingroup$ You don't know what it means for a function to be differentiable on $(-1,1)?$ $\endgroup$ – zhw. Nov 11 '16 at 19:08
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Hint for showing uniform convergence: If $a,b\ge 0,$ then $\sqrt {a+b} - \sqrt a \le \sqrt b.$

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Yes, it means that you have to prove that for any fixed $n$, $x \mapsto f_n(x)$ is differentiable (evident). Let $f(x) = |x|$.

To show that $(f_n)$ converges to $f$ in $C([-1,1])$, which is equipped with $d$, you have to show that $d(f_n,f) \to 0$.

$$|f_n(x) - |x| | = \sqrt{\frac1{n^2} + x^2} - \sqrt{x^2} = \frac{\frac{1}{n^2}}{\sqrt{\frac1{n^2} +x^2} + \sqrt{x^2}} \le \frac{\frac1{n^2}}{\frac1n} = \frac1n$$

This is true for all $x$, so:

$$d(f_n,f) = \sup_{x\in [-1,1]} |f_n(x) - |x|| \le \frac1n \to 0$$

i.e. $(f_n)$ converges to $f$ uniformly.

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$f_n$ is differentiable at$(-1,1)$ since

$\forall x\in (-1,1)\;\; \forall n\in \mathbb N \;\; \frac{1}{n^2}+x^2>0$

and

$$f'_n(x)=\frac{x}{ \sqrt{\frac{1}{n^2}+x^2} }.$$

for each $x\in (-1,1),$

$$ \lim_{n\to +\infty}f_n(x)=\sqrt{0+x^2}=|x|=f(x)$$

and

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

thus

$$d(f_n,f)\leq \frac{1}{n}$$

which proves that $(f_n)_n$ converges to $f$.

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