Let $v_\theta = (\cos\theta,\sin\theta)$ be a unit vector in the plane. I have a kernel $p(\theta,\theta') = p(v_\theta\cdot v_{\theta'})$ that satisfies $$\int_0^{2\pi} p(v_\theta\cdot v_{\theta'})\,d\theta' = 1\;\;\;(*)$$ for all $\theta\in [0,2\pi]$. I also have Chebyshev polynomials $T_0,T_1,\dots$ such that $T_k(cos\theta) = \cos(k\theta)$, normalized such that $$\{T_0/\sqrt{\pi}\}\cup\{\sqrt{2/\pi}T_k\}_{k=1}^\infty$$ form an orthonormal basis of $L^2(-1,1)$ with weight $1/\sqrt{1-t^2}$.
Now I write the Chebyshev expansion of my kernel: $$p(t) = \sum_{k=0}^\infty p_kT_k(t), \;\;\;\;\; t\in(-1,1)$$ and I want to show that $p_0 = \frac{1}{2\pi}$.
My progress so far: by orthonormality, we have \begin{align*} \int_0^{2\pi} &\frac{1}{\sqrt{\pi}}T_0(v_\theta\cdot v_{\theta'})\frac{\sqrt{2}}{\sqrt{\pi}}p(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1 - (v_\theta\cdot v_{\theta'})^2)}}\,d\theta'\\ &= \sum_{k=0}^\infty\int_0^{2\pi} T_0(v_\theta\cdot v_{\theta'})\frac{\sqrt{2}}{\sqrt{\pi}}p_kT_k(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1-(v_\theta\cdot v_{\theta'})^2}}\,d\theta'\\ &= \sqrt{2}p_0. \end{align*} Also, noting that $T_0\equiv 1$, I know \begin{align*} \int_0^{2\pi} &\frac{1}{\sqrt{\pi}}T_0(v_\theta\cdot v_{\theta'})\frac{\sqrt{2}}{\sqrt{\pi}}p(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1 - (v_\theta\cdot v_{\theta'})^2)}}\,d\theta'\\ &=\int_0^{2\pi} \frac{\sqrt{2}}{\pi} p(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1 - (v_\theta\cdot v_{\theta'})^2}}\,d\theta'. \end{align*} Then it would suffice to show $$\int_0^{2\pi} p(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1 - (v_\theta\cdot v_{\theta'})^2}}\,d\theta' = \frac{1}{2}.$$ This is where I'm stuck: I'm not sure how to use (*) in the expression above. Indeed since the expression above is constant in $v_\theta$ as we showed earlier, we are free to pick a particular value, say, $v_\theta = (0,1)$, to make this $$\int_0^{2\pi} \frac{p(\cos\theta')}{\sin\theta'}\,d\theta',$$ but still I am not sure what to do with this.
As it turns out, using orthonormality was a a red herring, and the solution is actually quite simple. Choosing $v_\theta = (1,0)$, we compute \begin{align*} 1 &= \int_0^{2\pi} p(v_\theta\cdot v_{\theta'})\,d\theta'\\ &= \sum_{k=0}^\infty \int_0^{2\pi} p_kT_k(v_\theta\cdot v_{\theta'})\,d\theta'\\ &= \sum_{k=0}^\infty \int_0^{2\pi} p_kT_k(\cos\theta')\,d\theta'\\ &= \sum_{k=0}^\infty \int_0^{2\pi} p_k\cos(k\theta')\,d\theta'\\ &= 2\pi p_0 + \sum_{k=1}^\infty \underbrace{\int_0^{2\pi} p_k\cos(k\theta')\,d\theta}_{= 0}, \end{align*} and so $p_0 = 1/2\pi$.