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Let us consider a sequence of real valued functions of real variable $x$, defined as
$f_n = g_n\;\;\;\;\;\;\;\;\;\;\;$ when $ a \leq x < b $
$f_n = g_n + \frac{1}{n}\;\;\;\;\;$ when $ b \leq x \leq c $
let each $g_n$ be continuous. Thus, each $f_n$ is piecewise continuous, with a "step" of height $\frac{1}{n}$ at $x=b$.
Let us now suppose that the sequence of the $g_n's$ is uniformly convergent on $[a, c] $. Then, also the $f_n's$ are uniformly convergent on $[a, c] $(said it roughly, because the height of the "steps" vanishes as $n \rightarrow \infty$).
This is a very simple example of a sequence of discontinuous functions (although piecewise continuous) uniformly converging to a limit function which is continuous on the whole of the domain $ a \leq x \leq c $.
In fact, one might even think of a similar sequence of functions, whereby also the position $b$ of the discontinuity is made to depend on $n$. It would still be converging uniformly to the same continuous limit function.

I have not had much success when trying to find literature on similar results and-or examples applicable to analytic complex valued functions of complex variable. This is for me interesting because in general, when switching to analytic functions, the corresponding theorems on uniform convergence becomes much more powerful.

I am particularly interested in the following question:

are there versions of Osgood and Vitali type of theorems for uniformly convergent "piecewise analytic" functions?

and also a different question, this related to uniform convergence on boundaries:

would a sequence of piecewise analytic functions that converges uniformly on the finite length boundary of a closed domain always converge uniformly to an analytic limit function in the domain itself?

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You should presuppose that the limit function is continuous at fisrt, otherwise, you can piece two "different" analytic functions together as the limit function. Suppose with the presupposition in mind. In my opinion, if there are a sequence of uniformly converging piecewise analytic functions which do not converge to an analytic limit , then Runge's Theorem should play an important role in the construction of the counterexample. – woodbass Dec 30 '12 at 12:28
"Piecewise constant" is an instance of "piecewise analytic", isn't it? And "continuous piecewise analytic" (with pieces having decent boundaries) is "analytic". So, what are really looking for? – fedja May 25 '13 at 13:28

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