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Let I(x) be a function such that I(x)=0 if x$\leq$0, and I(x)=1 if x$>0$. Let $c_{n}$ be a sequence of numbers such that $\sum$ c$_n$ converges absolutely. Let $x_n$ be some sequence of distinct real numbers.

Let $f_n(x)=\sum_{i=1}^{n} c_i I(x-x_i)$ and $f(x)=\sum_{i=1}^{\infty}c_iI(x-x_i)$.

I'm kind of stumped as to how to demonstrate both point-wise convergence and uniform convergence.

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up vote 3 down vote accepted

To show point-wise convergence, we must show that the sequence $\{f_n(x)=\sum_{i=1}^{n} c_i I(x-x_i)\}$ converges to $f(x)=\sum_{i=1}^{\infty}c_iI(x-x_i)$ for every value of $x$. This will follow trivially from the definition of an infinite series if we can show that the sequence converges at all. Since the space of real numbers is complete, it suffices to show that the sequence $\{f_n(x)=\sum_{i=1}^{n} c_i I(x-x_i)\}$ is Cauchy, that is for each $e > 0$ we can find an $N$ such that $\sum_{i=N}^{\infty} c_i I(x-x_i) < e$. But we can certainly find such an $N$ for $\sum_{i=N}^{\infty} c_i > \sum_{i=N}^{\infty} c_i I(x-x_i)$, thus we have that the sequence is point-wise convergent.

Uniform convergence, on the other hand, requires us to be able to choose $N$ independently of the value of $x$. However, we did not use the value of $x$ in selecting $N$ above, so uniform convergence follows as well.

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Hint: Ignore everything about the the way your function $I(x-x_i)$ is defined, except the fact that $|I(x-x_i)|\leq 1$ for all $x$. Everything then follows since the sum of the $c_i$ converges absolutely.

Let $C_n=\sum_{i=1}^n c_i$. Since $C_n$ is a Cauchy Sequence, we are able to conclude that $f_n (x)$ will be a Cauchy sequence at every $x$. But the fact that $f_n$ is Cauchy is independent of $x$ since it only depends on the convergence of $\sum c_i$, and hence it is uniformly convergent, as desired.

Hope that helps,

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