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Consider $T : C[0,1] \longrightarrow C[0,1]$ defined by $T(f(x)) = f^′(x)−f(x)$. I like this linear transformation because it's null space is functions of the form $ce^x$ for $c \in \mathbb{R}$ on the interval $[0,1]$. The range, on the other hand, I'm not quite as certain about. We are looking at all functions that can be written as a difference of a continuous function and it's derivative. Is this all of $C[0,1]$?

How can I convince myself of this?

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What is your definition of $C[0,1]$? – Aryabhata Mar 12 '12 at 7:37
Continuous functions on [0,1] – futurebird Mar 12 '12 at 7:41
So why is F'-F also continuous? – alpha.Debi Mar 12 '12 at 8:02
Every function in the domain must be differentiable, Not just contin. what is a name fir that domain? – futurebird Mar 12 '12 at 9:36
up vote 2 down vote accepted

Try to solve $$f'-f=g$$ This is a first order linear differential equation, that has a simple solution: $$f(x)=e^x\left(c+\int_0^x e^{-y} g(y) \, dy\right)$$ So you can get any $C[0,1]$ function as the difference between $f'$ and $f$.

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Why was it so hard to just look it that way when I already did the same thing with the null space? Thanks. – futurebird Mar 12 '12 at 7:42
Of course, I was a little too much of a physicist here. @Aryabhata is totally right. You can get almost any function and you should be careful about differentiability. But you can surely get all nice functions... – yohBS Mar 12 '12 at 13:24

Pick an $f$ whose derivative is not continuous everywhere (i.e $f'$ is not in $C[0,1]$). Then $f'(x) - f(x)$ cannot be in $C[0,1]$. A standard example is $f(x) = x^2 \sin(1/x)$ (with $f(0) = 0$).

Also, you talk of $T:C[0,1] \to C[0,1]$, but there are functions in $C[0,1]$ (based on your definition) which do not have a derivative (like the Weirstrass function) and the range, is a strict superset of $C[0,1]$.

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@alittledon: Or did I misunderstand your question? – Aryabhata Mar 12 '12 at 7:53
I think I need to ajust the domain... I recycled it from another problem. – futurebird Mar 12 '12 at 9:41

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