# What is the null vector for the vector space of continuous functions $f \colon \mathbf{R} \to \mathbf{C}$?

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 $$(f + g)(x) \equiv f(x) + g(x) \, ,$$ and multiplication of vector $\boldsymbol{f}$ by a scalar $a \in \mathbf{C}$ defined by $$(a f)(x) \equiv a f(x) \, ,$$ form a vector space.

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

## 1 Answer

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

• 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). Commented Nov 15, 2013 at 19:38
• @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.
– user61527
Commented Nov 15, 2013 at 20:30