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I studied existence of harmonic function on boundry by using Poission's formula. but some problem remain. Here is as follows

Suppose $Q(r,s)$ is a function on $\Bbb D =\{(r,\theta): 0 \le r \lt 1 , -\pi \le \theta \le \pi\}$ satisfying the following property.

(1) for any $r,\theta$ $Q(r,\theta) \ge 0$

(2) $\int_{-\pi}^{\pi} Q(r,t)dt=1$

(3) For any $\delta \gt 0$ $\lim_{r \to 1-} max\{Q(r,t) : \delta \le |t| \le \pi\}=0$

If $f$is a continuous function on the unit circle, define $v(r,\theta)= \int_{-\pi}^{\pi} f(\theta-t)Q(r,t)dt$. prove that $f_r(\theta)=v(r,\theta)$ converges to $f$ uniformly on $[-\pi,\pi]$

Attemp : Want to show

for some $r_0 \lt 1 $ such that $$|f(\theta)-f_r(\theta)| \le \epsilon$$ for all $r_0 \le r \lt 1 , -\pi \le \theta \pi$

since $f$ is unifomly continuous on closed set

For any $\epsilon \ge 0$ There is some $\delta \gt 0$ such that $|\theta -t| \le \delta$. $$|f(\theta)-f(t)| \le \epsilon$$

since above property (3). There is some $R \lt 0$

$$ max\{Q(r,t) : \delta \le |t| \le \pi\} \le \epsilon$$

for all $R\le r \le 1$

Now by using Poisson formula

$$f(\theta)-f_r(\theta) =\int_{-\pi}^{\pi}(\big(f(\theta) - f(\theta-t)\big)Q(r,t)dt$$

then split this integral in two

$$\le \int_{-\delta}^{\delta}(\big(f(\theta) - f(\theta-t)\big)Q(r,t)dt+\int_{-\pi}^{-\delta}+\int_{\delta}^{\pi}(\big(f(\theta) - f(\theta-t)\big)Q(r,t)dt$$ $$\le \epsilon \int_{-\delta}^{\delta}Q(r,t)dt +(-\delta +\pi)2M\epsilon+(\pi-\delta)2M\epsilon$$ $$\le \epsilon + 2M\pi\epsilon$$

where $M=\int_{-\pi}^{\pi}|f(t)|dt$

Thus for small $\epsilon$ we can make $$|f(\theta)-f_r(\theta)| \le \epsilon$$ QED

In above proof, I Know The property of (2),(3) for $Q(r,\theta)$ was used in this proof. but I think It can also be proved without using the condition of (1). where do the property be used in above proof? Thank you!

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  • If you think you can remove the positivity requirement, test your argument on the Dirichlet kernel for the Fourier series, and see if your argument would lead to uniform convergence of the Fourier series for continuous functions. – Disintegrating By Parts Nov 28 '17 at 20:04
  • According to your advice, I have to study about Dirichlet kernel. If I study completly, I will apply your advice. Thank you! – fivestar Nov 29 '17 at 06:31
  • You used the assumption in bounding $\int_{-\delta}^{\delta}((f(\theta)-f(\theta-t))Q(r,t)dt$, where you would otherwise need $|Q(r,t)|$. The Dirichlet kernel is oscillatory, and that's why you can't prove uniform convergence of the Fourier series for a continuous function. In fact, you can't prove pointwise convergence of the Fourier series for a continuous function. The Poisson kernel is much nicer in that regard. – Disintegrating By Parts Nov 29 '17 at 17:54
  • ahh.. I understand what you said vaguely... if there isn't first condition of $Q(r,t)$, it would need to be $\int_{-\delta}^{\delta}\big(f(\theta)-f(\theta-t)\big)|Q(r,t)|dt$, then we can't derive subsequent process as follows $\epsilon \int_{-\delta}^{\delta} Q(r,t) dt.$ thus the first condition is needed to prove uniform convergence. right? – fivestar Nov 30 '17 at 04:20
  • That's right. The Poison and Fejer kernels are positive, but the Dirichlet kernel is oscillatory, which kills pointwise and uniform convergence for continuous functions. – Disintegrating By Parts Nov 30 '17 at 04:21
  • @DisintegratingByParts Thanks for your kind explanation. It is helpful for me. and I also saw your post that explain some application about my question. It is difficult for me to understand slightly. but I am trying to do it repeatedly.Thanks your interests for me ! – fivestar Nov 30 '17 at 04:27

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