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I'm trying to derive the fundamental theorem of calculus myself and with as little help as possible. Is the following expression valid for all functions? I hope it is.$$f(a + b) = f(a) + f(b) % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWGHbGaey4kaSIaam % OyaiaacMcacqGH9aqpcaWGMbGaaiikaiaadggacaGGPaGaey4kaSIa % amOzaiaacIcacaWGIbGaaiykaaaa!3B49! $$

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    Is it valid for $f(x)=x^2$? – 2017-01-15
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    Or even for $f$ defined by $f(x)=1$. Basically, if you have a conjecture or question, try on the simplest examples first. – 2017-01-15
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    @Clement C. You are so right. I should have picked a simple function like that to answer my question, but I'm so frustrated deriving the theorem. I've been working tirelessly trying to solve it but can't and I'm pulling my hair out. – 2017-01-15
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    @MichaelLee If you are stuck (and don't want a full-fledged solution), you can also go on the Math.SE chat to interact with the people there? – 2017-01-15

4 Answers 4

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It is valid for linear functions, not for all functions.

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This formula is called Cauchy functional equation. In the class of continuous functions $f\colon\mathbb{R}\to\mathbb{R}$ the solutions are linear functions of the form $f(x)=xf(1)$. There are also discontinuous solutions (it was proved by Hamel in 1906). These ones are very strange. In particular, their graphs are dense subsets of a plane. Nevertheless, minimal regularity condition under $f$ fulfilling Cauchy equation makes it continuous (at every point) and - in consequence - linear. Such conditions are for instance boundedness on a neighbourhood of some point, measurability, lower (upper) semicontinuity at some point etc.

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No, that expression is not valid. Consider, for example, the function $f(x) = x^2$. We have $f(1 + 1) = f(2) = 2^2 = 4$, but $f(1) + f(1) = 1^2 + 1^2 = 2$.

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No, it's not.

For $f=\cos$ and $a=b=0$ you have $$\cos(a+b) = \cos(0) = 1 \ne 2 = 1+1 = \cos(0)+\cos(0) = \cos(a)+\cos(b)$$