# 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

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

• You forgot a square root on the integral defining the L2 distance. – Giuseppe Negro Mar 4 at 10:22

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 $$L^2$$.
• Thank you very much! However, isn't the limiting function discontinuous, since it explodes at 1/2? The exercise requires us to find a limiting function which is both continuous and periodic. The $f = \lim f_n$ is then even not a real valued function since $f(1/2) = \infty$. – Cebiş Mellim Mar 5 at 11:37
• @CebişMellim No, you are required to find a sequence $(f_n)_n$ converging in $L^2$ to some continuous periodic function $f$ but not pointwise. In my example, we have $f \equiv 0$. Then $f_n \to f$ in $L^2$ but not pointwise. – mechanodroid Mar 5 at 11:39