I've been trying to prove the following:

Lets have a function continuous and derivable for any x that belongs to the real numbers; and for any $x$ and $h: f(x+h) -f(x) = hf'(x)$.

So, I have to prove that $f(x)= ax+b$, where a and b are constants. I tried to use the Mean Value Theorem because it has its similarities but I kept getting nowhere. It is obvious that the function satisfies the equation but that ain't a proof. Thanks.

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    $\begingroup$ The writing can be better but this does not deserve the downvotes IMO. It's clear what is being asked and the OP shows some thoughts. $\endgroup$ Apr 23 '18 at 1:37
  • $\begingroup$ I agree. I don't usually answer downvoted questions, but this one seemed unfair. Also, I found more than one approach, so I like this question. $\endgroup$ Apr 23 '18 at 1:54
  • $\begingroup$ Since $f$ is differentiable you can differentiate the given relation with respect to $h$ and get $f'(x+h) =f'(x) $ for all $h$ and hence $f'$ is constant. Done!! $\endgroup$
    – Paramanand Singh
    Apr 23 '18 at 15:35

Note that $f(x+h) - f(x) = hf'(x)$ for all $x$ and $h$.

Fix any $y,j \in \mathbb R$. We want to show that $f'(y) = f'(y+j)$. If we show this, then $f'$ will be a constant, since it is equal at all values.

Let us highlight $x$ and $h$ in colour in the above premise. I am highlighting $x$ in green and $h$ in blue. $$ f(\color{green}x+ \color{blue}h) - f(\color{green}x) = \color{blue}h f'(\color{green} x) $$

Now, set $x = y$ and $y = j$. $$ f(\color{green}y+ \color{blue}j) - f(\color{green}y) = \color{blue}j f'(\color{green} y) \tag{*} $$

Now, set $x = y+j$ and $h = -j$. $$ f(\color{green}{y+j}+ \color{blue}{(-j)}) - f(\color{green}{y+j}) = \color{blue}{(-j)} f'(\color{green}{y+j}) $$ if you simplify the above, it becomes $$ f(y) - f(y+j) = (-j) f'(y+j) \tag{**} $$

Add the equations $(*)$ and $(**)$ to get: $$ 0 = j(f'(y) - f'(y+j)) $$

Now, if $j \neq 0$, then this forces $f'(y) = f'(y+j)$ for all $y, j \neq 0$ (of course, if $j = 0$ then $y = y+j$ so the equation is obvious). Hence, $f'$ is a constant.

Hence, for all $h \neq 0$, we have $hf'(x)=hf'(x+h)$, and then cancelling $h$ we get $f'(x+h) = f'(x)$. In conclusion, $f'$ is a constant function.

Now, if $f'(x) = a$ for all $x$, then if $f(0) = b$, you can easily verify that $f(x) = ax+b$ for all $x$, using the fact that $f(x+h) = f(x) + ha$ from the relation above.

Note : I'd like you to see if somehow we can show that $f$ is actually twice differentiable from the original premise, and that it's second derivative is identically zero. Then too we would be done.

  • $\begingroup$ Why can you switch $x$ and $x+h$ ? Thanks!. $\endgroup$
    – Karl
    Apr 23 '18 at 2:22
  • $\begingroup$ I want to answer this short, but I don't want to leave stones unturned. I will edit my answer and inform you when your question is answered. I request you to wait for five minutes. $\endgroup$ Apr 23 '18 at 2:56
  • $\begingroup$ Please see the edited answer. I request you to completely clarify the answer before accepting. $\endgroup$ Apr 23 '18 at 3:07
  • $\begingroup$ Please see the yellow box as well. I want to see if $f$ is twice differentiable from premise. $\endgroup$ Apr 23 '18 at 3:55

Clearly, $f'(x) = a$, the slope. Thus,

$$ f(x+h)-f(x) = a(x+h) + b - ax - b = ah = h \cdot a = h \cdot f'(x) $$


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