Counterexamples to equivalences between $L^n$-convergence types on periodic function spaces. I am working with $\mathbb{Z}$-periodic continuous functions from real to complex numbers, and I am comparing $L^2$, uniform and pointwise convergence. We have defined
$$\|f_n - f\|_{L^2} = \sqrt[2]{\int_{[0,1]} |f_n - f|^2}$$
$$\|f_n - f\|_{L^\infty} = \sup\{|f_n(x) - f(x)|: x\in\mathbb{R}\}$$
I have demonstrated, that uniform $L^{\infty}$ convergence implies $L^2$ convergence. 
In the following I have to give an example of a sequence $(f_n)_{n=1}^{\infty}$ of functions in $C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$ and another function $f \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$ such that


*

*$(f_n)_{n=1}^{\infty}$ converges to f in the $L^2$ metric, but does not converge to f uniformly.

*$(f_n)_{n=1}^{\infty}$ converges to f in the $L^2$ metric, but does not
converge to f pointwise.

*$(f_n)_{n=1}^{\infty}$ converges to f pointwise, but does not converge to f in the $L^2$ metric.


For 1. I had the 1-periodic sine function in mind. The sine function $\sin(2\pi x)^n$ goes to zero everywhere but at $x=0.25,0.75$. Thus, its measure goes to zero with large n, but its supremum stays at 1. However, I do not know how to show rigorously that $\int_{[0,1]} |\sin(2\pi x)^n|^2 dx \leq \epsilon$. Also, I just realised that its limit is not in the space, since it is not continuous.
For 3. I would take $1/x$, but I do not know how to make it periodic and continuous.
Any help is greatly appreciated!
 A: First notice that you are pretty much working on the space
$$X = \{f \in C[0,1] : f(0) = f(1)\}$$
because every function from $X$ can be extended by periodicity to a continuous $1$-periodic function $\mathbb{R} \to \mathbb{R}$, and conversely, the restriction of a continuous $1$-periodic function $\mathbb{R} \to \mathbb{R}$ to $[0,1]$ belongs to $X$.
For $(a)$ and $(b)$ consider the sequence
$$f_n(x) = \begin{cases} \sqrt{n^3x - \frac{n^3}2 + n},& \text{if } x \in \left[\frac12 - \frac1{n^2}, \frac12\right]\\
\sqrt{-n^3x + \frac{n^3}2 + n},& \text{if } x \in \left[\frac12, \frac12 + \frac1{n^2}\right]\\
0,& \text{otherwise}\end{cases}$$
Then $f_n \xrightarrow{L^2} 0$ because the graph of $|f_n|^2$ is an isosceles triangle centered at $\frac12$ with base length $\frac2{n^2}$ and height $n$. However, we have $f_n\left(\frac12\right) = \sqrt{n}$ which is unbounded so $(f_n)_n$ doesn't converge pointwise or uniformly.
For $(c)$ similarly consider
$$g_n(x) = \begin{cases} \sqrt{n^2x - n},& \text{if } x \in \left[\frac1{n}, \frac2n\right]\\
\sqrt{-n^2x +3n},& \text{if } x \in \left[\frac2n, \frac3{n}\right]\\
0,& \text{otherwise}\end{cases}$$
The graph of $|g_n|^2$ is an isosceles triangle centered at $\frac2n$ with base length $\frac2{n}$ and height $n$. Hence $g_n$ is supported on $\left[\frac1n, \frac3n\right]$ so $g_n \to 0$ pointwise. Hence the only candidate for $L^2$ convergence is $0$ but we have $\|g_n\|_2 = 1$ so $(g_n)_n$ doesn't converge in $L^2$.
