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$f : \mathbb{R} \to \mathbb{R}$ is a differentiable function satisfying $$f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}$$

if $f'(0)=2$, find the function

My Try:

we have $$f\left(\frac{x+y}{3}\right)-\frac{f(y)}{3}=\frac{2+f(x)}{3}$$ $\implies$

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

Now taking Limit $x \to 0$ we have

$$\lim_{x \to 0}\frac{f\left(\frac{x+y}{3}\right)-\frac{f(y)}{3}}{\frac{x}{3}}=\lim_{x \to 0}\frac{2+f(x)}{x}$$

$\implies$

$$f'\left(\frac{y}{3}\right)=\lim_{x \to 0}\frac{2+f(x)}{x}$$

Now since LHS to be finite , we need $0/0$ form in RHS, hence $f(0)=-2$

Now by L'Hopital's Rule we get

$$f'\left(\frac{y}{3}\right)=f'(0)=2$$

Integrating we get

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

Putting $y=0$ we get $c=-6$

So

$$3f\left(\frac{y}{3}\right)=2y-6$$

So $$f(y)=2y-2$$

Hence $$f(x)=2x-2$$

But this function is not satisfying given functional equation.

What went wrong?

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  • 1
    $\begingroup$ The limit which you claimed to be $f'(y/3)$ is not, except perhaps if you implicitly assume the linearity of the map $f$, which would be a contradiction with its defining equation. By the way, setting $x=y=0$ in the defining equation, one gets $f(0) = 2$. In fact, the affine map $a(x) = 2x+2$ solves the equation for $f$ and the derivative constraint. $\endgroup$ – Jordan Payette Jan 25 '18 at 3:31
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Using $x=y=0$ in the functional equation we can see that $f(0)=2$. Further note that $$f(x+h)=\frac{2+f(3x)+f(3h)}{3}, f(x) =\frac{2+f(3x)+f(0)}{3}$$ so that $$\frac{f(x+h) - f(x)} {h} =\frac{f(3h)-f(0)}{3h}$$ Taking limits as $h\to 0$ we get $f'(x) =f'(0)=2$ and thus $f(x) =2x+c$ and from $f(0)=2$ we get $c=2$ so that $f(x) =2x+2$. You can check that it satisfies the functional equation.

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One readily verifies that the affine map $a(x) = 2x+2$ is a differentiable function which solves both the defining equation for $f$ and the derivative constraint $f'(0) = 2$.

Let's prove now the uniqueness of this solution. The defining equation can be rewritten $f\left( \frac{x+y}{3} \right) - \frac{f(x) + f(y)}{3} = \frac{2}{3}$. Differentiating with respect to $x$, we get $f'\left( \frac{x+y}{3} \right) - f'(x)=0$ regardless of the values $x$ and $y$, so that $f'(x)$ is a constant. The constraint $f'(0)=2$ thus yields $f'(x) \equiv 2$, which integrates to $f(x) = 2x+c$. Substituting this result into the defining equation for $f$ readily yields $c=2$.

Remark: Paramanand Singh's answer is better than mine in that it proves that $f$ is differentiable at every point by only using its defining equation and the assumption that it is differentiable at $0$. In that respect it is quite instructive. Be aware however that such a proof is in general rather difficult, hence the frequent assumption that $f$ is differentiable ; one doesn't need then to rely on the definition of a derivative, but can rather merely differentiate the defining equation as I did above.

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