Is T a Linear Transformation from V to R1?

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SUMMARY

The discussion confirms that the operator T, defined as T(f) = ∫ from 0 to 1 f(x) dx for f in V (the space of all continuous functions on [0,1]), is indeed a linear transformation from V to R1. The proof demonstrates linearity by showing that T(f + g) = Tf + Tg and T(kf) = kTf for any functions f, g in V and scalar k. The conclusion is that T satisfies the properties required for linear transformations, confirming its classification as such.

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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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