Sequence of continuous functions which converges to a continuous limit Any help with this: construct a sequence of continuous functions defined on $ [0,1] $ which converges pointwise but not uniformly to a continuous limit ?
Thank you.
 A: Hint: Consider $f_n: [0,1] \to \Bbb R$ defined by 
$$f_n(x) := \begin{cases}
 2n^2x   & \text{if $0 \le x \le \frac1{2n}$}\\
 2n-2n^2x & \text{if $\frac1{2n} \le x \le \frac 1n$}\\
 0              & \text{if $x \ge \frac1n$}
\end{cases}$$
What is the pointwise limit of the $f_n$?
A: Consider $f_n: [0,1] \to \Bbb R$ defined by 
$$f_n(x) := \begin{cases}
 nx   & \text{if $0 \le x \le \frac1{n}$}\\
 2-nx & \text{if $\frac1{n} \le x \le \frac 2n$}\\
 0              & \text{if $ x \ge \frac2n $}
\end{cases}$$
It is not hard to verify that the pointwise lmit of $f_n(x)$ is 0. In fact, it is obvious that
. Now for any $x \in (0,1]$ we will have a sufficiently large N such that
$f_n(x) = 0$ if $n \geq  N$. However, $f_n(x)$ does not converge uniformly to 0 due to the fact that $\sup|f_n(x) - f(x)| = 1$ for $x\in [0,1]$
A: Consider $f_n$ the linear interpolation of $(0,0),(n^{-1},1),(2n^{-1},0),(1,0)$. 
A: Take a sequence of null functions that have a triangular jump of length $1/n$ and height $n$ on $x=1/2n$.
A: Hint:  to have it converge pointwise but not uniformly you need the function to get steeper and steeper somewhere.  A natural try would be $x^n$ which gets steeper and steeper near $1$ as $n \to \infty$.  Unfortunately, the resulting function is not continuous:  it is $0$ for $x \in [0,1)$ but $1$ for $x=1$  Can you see how to fix this?
