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Let $f_n:[0,1]\rightarrow \mathbb{R}$ be defined by $f_n(x)=\frac{n}{2}x^2$ if $0\leq x\leq \frac{1}{n}$ and $f_n(x)=x-\frac{1}{2n}$ if $ \frac{1}{n}<x\leq 1.$

Show that:

1) $f_n$ is continuously differentiable on $[0,1]$ for all $ n=1,2,\cdots $

2) $(f_{n})_{n\geq 1}$ is uniformly convergent on $[0,1]$.

3) $(f'_{n})_{n\geq 1}$ is point-wise convergent, but not uniformly convergent on $[0,1].$

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What have you tried so far? –  oks Feb 13 '13 at 13:50

1 Answer 1

HINTS:

$1)$ the only problem could occur at the gluing point $x = 1/n$. Compute the derivatives from both sides to show that it doesn't.

$2)$ imagine $n$ is very big. Then the graph of $f_n$ looks almost like a straight line $f(x) = x$. Try to prove that $f(x) - f_n(x)$ indeed goes to zero uniformly.

$3)$ It almost holds that $f_n'(x) = 1$ if it weren't for the initial interval $[0,1/n)$ where the derivative is close to zero. Show, that no matter how big $n$ gets, the derivative will always be close to zero on some initial interval.

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