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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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