Sequence of continuous functions which converges to a continuous limit

Question

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.

Answer 1

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 $x\in [0,1]$

Answer 2

Consider $f_n$ the linear interpolation of $(0,0),(n^{-1},1),(2n^{-1},0),(1,0)$.

Answer 3

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 $f_n$?

Answer 4

Take a sequence of null functions that have a triangular jump of length $1/n$ and height $n$ on $x=1/2n$.

Answer 5

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?