# Extending an operator from $C([0,1])$ to $L^2([0,1])$

Mathematics Asked by AspiringMathematician on October 7, 2020

Suppose I have a "time-sampling" operator given by
begin{align*} S_m: C([0,1]) &to mathbb{R}^m \ f &mapsto (f(t_1),f(t_2),…,f(t_m)) end{align*}

Now I want to extend this to $$L^2([0,1])$$. However, what would happen to the operator if the function was discontinuous exactly at the points $$t_i$$? For example, consider some piecewise-constant function $$f$$, whose discontinuities lie exactly at the points $$t_i$$. Is there a natural way to define what should $$f(t_i)$$ be?

If such an extension is impossible, then my "gut feeling" says that, at least for piecewise-continuous functions (which is the main case I’m considering), if $$f$$ has discontinuities at $$t_i$$, then I should take
$$begin{equation*} f(t_i) = frac{f^+(t_i)+f^-(t_i)}{2} end{equation*}$$
where $$f^+(t_i)$$ and $$f^-(t_i)$$ are the one-sided limits. This feeling is based on Fourier series behavior at such discontinuities. But that’s it, at most a feeling.

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