an example for spherical harmonic

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Let $S=\{z\in \mathbb{C}: |z|=1\}$ be the circle of $\mathbb{C}$. We consider the applications $h^{p}(z)$ defined on $\Bbb{C}\ni z\to\mathbb{C}$ with $(p\in \mathbb{N}^*)$ such that :

  1. $h^{p}(rw)=r^ph^{p}(w),\quad r>0$ and $\int_{S}h^{p}(w)d\sigma(w)=0$ where $d\sigma(w)$ is the Lebesgue measure on $S$.

  2. $\int_{S}|h^{p}(w)|^2d\sigma(w)=1$ and $\int_{S} h^{p}(w)\overline{h^{q}(w)}d\sigma(w)=\delta_{p,q}$.

Put $\Gamma:=\{ g(w)=\sum_{p\geq0} a^ph^p(w) (w\in S): ||g||^2=\int_{S} |g(w)|^2d\sigma(w)=\sum^{\infty}_{p\geq0}|a^p|^2<\infty \}$ with $a_p\in \mathbb{C}$

By using 1 et 2 on can shows: $\Gamma:=\{ g(w)=\sum_{p\geq0} a^ph^p(w) (w\in S): \sum^{\infty}_{p\geq0}|a^p|^2<\infty \}$ with $a_p\in \mathbb{C}$.

Now, put $\Gamma_r=\{g\in \Gamma: \int_{S^1}|g(w)|^rd\sigma(w) <\infty\},\quad r>2$ so $\Gamma_r\subset \Gamma$.

My question is. I m looking for an elemnt $g\in \Gamma$ such that $g\not\in \Gamma_r$.

Thank you a lot of.