Complex analysis about Cauchy's integral formula Let g(t) be any continuous function on the boundary T of Disk D(0,1) centered at 0 and radius 1.
Then, extend g(t) to the interior of the disc by $g(z) = \int_T g(t)/(t-z) dt $
If it is well defined, my question is:
(1), is g(z) continuous on the whole closed disk.
(2, can g(z) always be able to extended a little bit to a bigger open disk so that g(z) is holomorphic on that disk.
 A: This is more of an extended comment, since it's not a definitive answer to either questions.
Notice that expanding in a power series we have
$$
\int_T \dfrac{g(\zeta)}{\zeta-z}\, d\zeta= \int_T \sum_{k=0}^\infty \zeta^{-1}\left( \dfrac{z}{\zeta}\right)^k g(\zeta) \, d\zeta= \sum_{k=0}^\infty z^k \int_T \zeta^{-k-1}g(\zeta)\, d\zeta,
$$
where the change of sum an integral is justified by the uniform convergence of the series on compact subsets of $|z|<1$. Now parametrize $T$ counterclockwise by $\zeta=e^{2\pi i t}$, with $0\leq t\leq 1$ (so that $d\zeta/\zeta= 2\pi i dt$) to write
$$
\int_T \dfrac{g(\zeta)}{\zeta-z}\, d\zeta = 2\pi i\sum_{k=1}^\infty z^k \int_0^1 e^{-2\pi ikt}g(t)\, dt= 2\pi i\sum_{k=0}^\infty z^k \hat{g}(k),
$$
where we have abused notation a little and identified $g(e^{2\pi it})$ and $g(t)$, and $\hat{g}(k)$ denotes the $k$-th Fourier coefficient of $g$.
If you write $z=re^{2\pi i \theta}$, then the above reads
$$
\dfrac{1}{2\pi i} \int_T\dfrac{g(\zeta)}{\zeta-z}\, d\zeta = \sum_{k=0}^\infty r^k e^{2\pi i k\theta}\hat{g}(k).
$$
So as $r\to 1^-$, we see that the limit should be the function $\tilde{g}$, built from $g$ by deleting all the negative Fourier coefficients. You can get several types of convergence here (as $r\to 1^-$): We have convergence in $L^2$, nontangential convergence (meaning approaching boundary points along cones inside the disk).
However I'm not sure at the moment if you can even conclude that the limit function $\tilde{g}$ needs to be continuous (if you want to look more into this, look up Hardy spaces of the unit disk), so as of now I don't have a definite answer to (1) or (2) beyond pointing out what the limit is.
On the plus side (at least for (1)), I'd imagine that a Littlewood-Paley characterization of Hölder continuous functions should allow you to conclude in the affirmative for (1), when $g$ is in this class.
Added: A negative answer to (2) can be given as follows. Consider the function $g$ whose Fourier coefficients are given by
$$
\hat{g}(k)=
\begin{cases}
\frac{1}{k^2} & k>0,\\
0 & k\leq 0.
\end{cases}
$$
Such a $g$ is continuous, owing to the fact that it's Fourier series is absolutely convergent. On the other hand, we know that its extension to the unit disk is given by
$$
\sum_{k=1}^\infty \dfrac{1}{k^2}z^k= \sum_{k=1}^\infty a_k z^k,
$$
so that its radius of convergence of this power series is 1 (since $\limsup |a_k|^{1/k}=1$).
