Is T a Linear Transformation from V to R1?

AI Thread Summary
The discussion centers on proving that the transformation T, defined as the integral of a continuous function f over the interval [0,1], is a linear transformation from the space of continuous functions V to the real numbers R1. The proof demonstrates linearity by showing that T(f + g) equals Tf + Tg and T(kf) equals kTf, confirming that T satisfies the properties of linearity. The poster seeks validation for their proof and expresses concern about potentially missing nuances in the definition of the transformation. A response clarifies that the notation f: V → R indicates that the integral of a function in V results in a real number. Overall, the proof is affirmed as correct and straightforward.
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I'd like to check my proof. It seems easy enough, but I'd like to make sure that I'm not missing anything:

If V is the space of all continuous functions on [0,1] and if
Tf = integral of f(x) from 0 to 1 for f in V, show that T is a linear transformation From V into R1.

Like I said the proof seems simple enough, but I just want to make sure I'm not missing anything that might be implied by "From V into R1."

T(f + g) = integral from 0 to 1[f(x) + g(x)]dx
= integral from 0 to 1 f(x)dx + integral from 0 to 1 g(x)dx
= Tf + Tg

T(kf) = integral from 0 to 1 kf(x)dx
= k*integral from 0 to1 f(x)dx
= kTf

there for T is a linear transformation.

I feel silly posting something this simple, but I'm just not absolutely sure that I'm not missing something.

Thanks as usual for all the help.
 
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Yeah, that looks fine.

f:V \rightarrow\ R just refers to the fact that the definite integral over a function in V will always give you a constant real number.
 
Thanks Stevo.
 
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