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Given a sequence of polygonal arcs $(f_n)_{n\in\mathbb{N}}$ that has limit $f$, is $f$ continuous?

I believe it should be true as I cannot think of a counter example but not sure how to prove it.

Thank you for your help.

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    what is a polygonal arc. In case of functions this is not true. Consider $f_{n}(x)=x^{n}$ on $[0,1]$.2011-07-31
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    If a polygonal arc is what I think it is, you could for instance take $n$ points along $f_n$ from Chandru's example to get a sequence of functions defined by polygonal arcs that converges to a discontinuous function.2011-07-31
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    I'm guessing a polygonal arc can be interpreted as a continuous piecewise linear function. Perhaps if $\lim f_n$ exists it's continuous almost everywhere? I don't know enough analysis to say.2011-07-31
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    With $x^n$, there is a discontinuity at $1$. But the limiting *curve* is nice.2011-07-31
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    @Chandrasekhar: but your sequence converges to $0$ in $[0,1]$, and the constant functions are continuous. What am I missing?2011-08-28

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Let $f_1$ be the polygonal arc defined thus. Let $P_0=(1,0)$, $P_1=(2/3,1)$, $P_2=(1/3,0)$, $P_\infty=(0,0)$. Join $P_0$ to $P_1$ to $P_2$ to $P_\infty$ by straight line segments.

Let $f_2$ be the polygonal arc defined thus. Let $P_3=(2/9,1)$, $P_4=(1/9,0)$. Join $P_0$ to $P_1$ to $P_2$ to $P_3$ to $P_4$ to $P_\infty$ by straight line segments.

In general, for $k \ge 1$, let $P_{2k-1} =(2/3^k,1)$ and $P_{2k}=(1/3^k,0)$, and let $f_n$ be the polygonal arc obtained by joining $P_i$ to $P_{i+1}$, for $i=0,1,\dots, 2n-1$, and finally $P_{2n}$ to $P_\infty$, by straight line segments.

We have imitated the behaviour of $\sin^2(1/x)$. The limiting curve is not continuous at $(0,0)$. If an endpoint difficulty is not acceptable, we can modify the definitions by adding to each polygonal arc $f_n$ its reflection across the $y$-axis.