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Suppose a continuous ,differentiable and real valued function satisfies the relation $f(\frac{x+y}{3})=\frac{2+f(x)+f(y)}{3}$ For all real x and y. If $f’(2)=2$ Then find $f(x)$

This is how I attempted it:

We substitute $y=2x$ in the equation $f(\frac{x+y}{3})=\frac{2+f(x)+f(y)}{3}$ we get , $f(x)=\frac{2+f(x)+f(2x)}{3}$. Now , rearranging the equation gives us ,

$2f(x)=2+f(2x)$

Differentiating both sides ,

$2f’(x)=2f’(2x)$ or $f’(x)=f’(2x)$

Now if we substitute $x=1,2,3,$ etc., we will get $f’(x)=2$ for all $x$ . Therefore I concluded ,

$f’(x)=2$ for all $x$

Now integrating both sides ,

$f(x)=2x+c$ , where $c$ is the constant of integration. By substituting $x=0$ in the original functional equation , we get $f(0)=2$ this leads us to the conclusion that $f(x)=2x+2$

All was well till here and the answer is also right. However after sometime , I suddenly realized that I had only considered multiples of two. So I presume that this equation will work only for multiples of two. Then it means that this wouldn’t work for numbers like $3$ or $5$ . Does this mean that my method is wrong ? Or a rational explanation to my confusion would be that odd numbers like $3$ or $5$ can be written as $2(3/2)$ or $2(5/2)$ etc.? Do you think that the method I’ve used is correct ? Please suggest mistakes if any . Thanks !

Aditi
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  • If we can get a real value for $f$ after substituting a particular value $x_0$ then we say the function is working for the particular value $x_0$. Then what's wrong with substituting odd numbers? – Extremal Jul 29 '18 at 16:47
  • @Extremal Well I substituted even numbers because I got $f’(2x)=f’(x)$ and I had the value of $f’(2)$ so it was easy to get the values of $f’(4),f’(8)$ but I can’t figure out how to get the value of $f’(3)$ . – Aditi Jul 29 '18 at 16:50
  • The argument is not correct, and you didn't substitute multiples of $2$, you substituted powers of $2$. –  Jul 29 '18 at 16:51
  • @nextpuzzle , so is the argument not correct because it wouldn’t work for numbers like $3,6$ etc ? – Aditi Jul 29 '18 at 16:52
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    You can take derivative with respect to $x$ on the original equation to get $\frac{f'((x+y)/3)}{3}=\frac{f'(x)}{3}$. Since this equation is true for all $y$, then $f'$ is constant. Since $f'(2)=2$, then $f'(x)=2$ for all $x$. –  Jul 29 '18 at 16:52
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    @Aditi The argument can be made work. I am only saying that as written now, it is wrong. –  Jul 29 '18 at 16:53
  • @nextpuzzle if I wanted to correct this argument , could you recommend a suggestion? Or is it not possible to do it by this method ? – Aditi Jul 29 '18 at 16:54
  • @Extremal I had assumed that $f’(x)=2$ for all $x$ before substituting powers of two. Then I realized I hadn’t substituted other numbers – Aditi Jul 29 '18 at 16:55
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    why not just plug in $f(x)=2x+2$ into $f(\frac{x+y}{3})=\frac{2+f(x)+f(y)}{3}$, the result is an identity in $x$ and $y$ – WW1 Jul 29 '18 at 17:01
  • @WW1 thanks for the suggestion. I will try that as well. I just wanted to correct the method that I’ve used anyway. But now I realize that I’m probably wrong. – Aditi Jul 29 '18 at 17:03
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    @WW1 Existence does not imply uniqueness. – Mark Viola Jul 29 '18 at 17:16

1 Answers1

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From the functional equation

$$f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}\tag1$$

it is trivial to see that

$$f(0)=2\tag2$$

Differentiating $(1)$ with respect to $y$ reveals

$$f'\left(\frac{x+y}{3}\right)=f'(y) \tag3$$

whereupon setting $y=2$, we find that for all $x$

$$f'\left(\frac13 x+\frac23\right)=2\tag4$$

Integrating $(4)$ and applying $(2)$ yields the coveted relationship

$$f(x)=2x+2$$

Mark Viola
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