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The set of all continuous complex-valued functions of real variable $x$ together with addition of two vectors $\boldsymbol{f} = f(x)$ and $\boldsymbol{g} = g(x)$ defined by \begin{equation} (f + g)(x) \equiv f(x) + g(x) \, , \end{equation} and multiplication of vector $\boldsymbol{f}$ by a scalar $a \in \mathbf{C}$ defined by \begin{equation} (a f)(x) \equiv a f(x) \, , \end{equation} form a vector space.

What is the null vector $\boldsymbol{0}$ in this space? It it defined by $\forall x \in \mathbf{R} \colon 0(x) \equiv 0 \, ?$

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Yes, you're correct: It is the zero function that always returns $0$ for any input.

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  • $\begingroup$ Just a quick follow-up that came into my head right now: strictly speaking is it also true for $L^2$? Or there is something funny with the "Lebesgue measure zero" as in the case of the equivalence of two functions? I don't know what this "Lebesgue measure zero" actually is, but I just read that two functions on $L^2$ are considered to be equivalent if the are equal everywhere except on a subset of the real numbers that have this "Lebesgue measure zero" (whatever it is). $\endgroup$
    – Wildcat
    Commented Nov 15, 2013 at 19:38
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    $\begingroup$ @Wildcat The elements of $L^2$ are really equivalence classes of (pointwise-defined) functions, so the zero element is the equivalence class of the function which is zero everywhere. $\endgroup$
    – user61527
    Commented Nov 15, 2013 at 20:30

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