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Let $f$ be a differentiable function such that for every $x$ and $h$ it holds that $f(x+h)-f(x)=hf'(x)$. Prove that $f(x)=kx+n$ where $k$ and $n$ are constants.

I get it why this is true, and I tried to prove it somehow, but I can't seem to prove it rigorously with analysis. Any ideas?

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    $\begingroup$ Nitpick: this kind of function is usually called affine. $\endgroup$
    – Sasho Nikolov
    Commented Sep 27, 2016 at 22:14
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    $\begingroup$ @SashoNikolov Indeed. Linear is something like that $f(x + y) = f(x) + f(y)$ and $f(kx) = kf(x)$. $\endgroup$
    – Kaz
    Commented Sep 28, 2016 at 0:42
  • $\begingroup$ @Kaz It depends on the discipline. Linear equations are often taken to mean polynomials of degree $1$ which need not fix $0$ $\endgroup$
    – Henry
    Commented Sep 28, 2016 at 8:13

2 Answers 2

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The given equation $f(x+h)−f(x)=hf′(x)$ holds for all real $h, x$, so we can safely set $x=0$. This gives $$f(h)-f(0)=hf'(0)\iff f(h)=f'(0)\cdot h+f(0).$$ Letting $h=x,f'(0)=k,f(0)=n$, we have $f(x)=kx+n$ as desired. $\blacksquare$

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Differentiate with respect to $h$. You get

$$f'(x+h)=f'(x)$$ for every $h$ and $x$, therefore the derivative is constant.

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