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I was looking at Cauchy's functional equation when I happened upon an interesting result.

Consider a differentiable function $f:\mathbb R\to\mathbb R$ with $f(1)=1$ and $f(x+y)=f(x)+f(y)$.

With $x=y=0$, it follows that $f(0)=0$.

...and then I do something not so orthodox:

$$f(x+h)=f(x)+f(h)$$

$$f(x+h)-f(x)=f(h)=f(0+h)-f(0)$$

$$\frac{f(x+h)-f(x)}h=\frac{f(0+h)-f(0)}h$$

as $h\to0$, we end up with, nicely,

$$f'(x)=f'(0)$$

Thus, $f(x)$ is linear, and interpolating with $f(0)=0$ and $f(1)=1$, we end up with

$$f(x)=x$$

just as suspected.

Now, I don't usually see this done with functional equations, using derivatives and all, so I was wondering 2 questions:

  1. Is what I did ok, given the conditions on $f$?

  2. Could someone produce an interesting scenario of when a hard to solve functional equation reduces nicely with derivatives? Thanks :-)

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    Not answering your question, but from a more algebraic perspective I think it is sufficient for $f$ to be continuous, since your conditions force $f$ to be the identity on $\mathbb{Q}$. – 2017-01-23
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    Of course what you did is okay, since you assumed differentiability. It would have sufficed to assume differentiability at $0$ (or at any single point) and the functional equation to obtain the same result, without any explicit regularity assumptions anywhere else. – 2017-01-23

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Here is an interesting example of a functional equation that can be solved more easily using derivatives: $$f(x+1)-f(1-x)+f(2x)=x^2+x+1$$ By taking the derivative of both sides, we have $$f'(x+1)+f'(1-x)+2f'(2x)=2x+1$$ and by differentiating again, $$f''(x+1)-f''(1-x)+4f''(2x)=2$$ At this point it seems safe to assume that $f$ is constant, or that $f(x)=a$ for some constant $a$. Then we have $$a-a+4a=2$$ $$a=\frac{1}{2}$$ and so $$f''(x)=\frac{1}{2}$$ $$f'(x)=\frac{1}{2}x+b$$ for some other constant $c$. Then we can discover from the equation $$f'(x+1)+f'(1-x)+2f'(2x)=2x+1$$ that $$\frac{1}{2}+b+\frac{1}{2}+b+2b=1$$ $$4b+1=1$$ $$b=0$$ and so we have $$f'(x)=\frac{1}{2}x$$ and $$f(x)=\frac{1}{4}x^2+c$$ for some constant $c$. Finally, from the original functional equation $$f(x+1)-f(1-x)+f(2x)=x^2+x+1$$ we have $$\frac{1}{4}+c-\frac{1}{4}-c+c=1$$ $$c=1$$ And, at long last, we have $$\color{red}{f(x)=\frac{1}{4}x^2+1}$$ Most polynomial functional equations can be solved easily this way. A solution can be obtained to the original equation by assuming that $f$ was quadratic and solving for $a$, $b$, and $c$, but that would involve a lot of messy algebra, and a neatly partitioned answer involving derivatives is highly preferable.

More examples like this can be found here.