Finding a pointwise convergent sequence that does not converge in the square mean

Question

I am having a tough time finding a function sequence $(f_{n})_{n\in\mathbb N}$ of continuous functions of $[0,1] \to \mathbb K$, whereby $\mathbb K \in \{\mathbb R,\mathbb C\}$, such that $(f_{n})_{n\in\mathbb N}$ converges pointwise but does not converge in quadratic mean.

My ideas: Since $\forall x \in [0,1]: |f_{n}(x)-f(x)|\to 0, n\to \infty.$

Basically I'm trying to find a $C>0$, such that $\int^{1}_{0}|(f_{n}-f)(x)|^{2}dx\geq\int^{1}_{0}C^{2}dx \neq0. $

Any hints for suited functions?

Answer 1

Let $$f_n = \begin{cases} \sqrt{n\pi\sin(n\pi x)} \text{ for } 0 \leq x \leq \dfrac{1}{n} \\ 0 \text{ for } \dfrac{1}{n} < x \leq 1 \end{cases} $$

Then $\{f_n\}$ converges pointwise to $0$. To see this, note that for any $c \in (0, 1]$, then for all $k > \frac{1}{c}$,

$$f_k(c) = 0$$

Moreover, $f_k(0)$ is always $0$.

However, $\{f_n\}$ does not converge to $0$ on the quadratic mean. It is easy to see that

$$\int_{0}^{1} f_k^2(x) \, dx = 2 \neq 0$$

always.

For visual intuition, view here.

Answer 2

The sequence $f_n(x) = n\sqrt {x^n(1-x)}$ converges pointwise to $0$ everywhere on $[0,1],$ but $\int |f_n-0|^2 \to 1.$