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Find all function $f,g$ that satisfy: $$g(x)-g(y)=\frac{1}{6} (x-y)(f(x)+f((x+y)/2)+f(y))$$ For $y=0$ we have an equation in $f$: $$4(x-y)(f(x/2)-f(x/2+y/2))=xf(y)-yf(x)$$ How can i do it?

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Could you kindly typeset your question to make it more readable?… – user17762 Nov 22 '12 at 21:38
Why does $y$ appear in the equation with $y=0$ and why does $g$ not appear? – Hagen von Eitzen Nov 22 '12 at 22:03

2 Answers 2

No continuity assumption is needed. The final solution is:

$$f(x)=ax+b, \quad g(x)=\frac{1}{4}ax^2+\frac{1}{2}bx+c$$

where $a, b, c$ are constants.

Warning: Long solution ahead.

Step 1: Determine $g$ in terms of $f$, so we can focus on an equation with only $f$.

Let $g(0)=c$. Put $y=0$, so $$g(x)-c=\frac{1}{6}x(f(x)+f(\frac{x}{2})+f(0))$$

Thus $g(x)=c+\frac{1}{6}x(f(x)+f(\frac{x}{2})+f(0))$.

Substituting this back into the original equation gives


Some rearranging gives


Step 2: A simplification to remove $f(0)$.

Define $h(x)=f(x)-f(0)$, so $h(0)=0$.

Then the equation becomes


which we denote as $P(x, y)$.

Step 3: Proving that $h$ is an odd function.

$P(x, -x)$ gives



For $x \not =0$, we thus have


Since this clearly holds for $x=0$ as well, we have


for all $x$.

Define $k(x)=h(x)+h(-x)$. Thus $k(x)=k(\frac{x}{2})$, and $k(0)=2h(0)=0$.

Now $P(-x, -y)$ gives



Adding this to $P(x, y)$ gives


Since $k(x)=k(\frac{x}{2})$ this gives



Put $y=2x$ and use $k(2x)=k(x)$ to get


Replace $x$ by $\frac{x}{3}$, so $-\frac{x}{3}k(x)=0$. Thus for $x \not =0$, $k(x)=0$. Since $k(0)=0$, we have $k(x)=0 \, \forall x$.

Thus $0=k(x)=h(x)+h(-x)$ so indeed $h$ is an odd function.

Step 4: Proving $h(2x)=2h(x)$.

$P(x, -2x)$ gives




Thus for $x \not =0$, we have $4h(\frac{x}{2})=4h(x)-h(2x)$. Since this holds for $x=0$ as well ($h(0)=0$) we have

$$4h(\frac{x}{2})=4h(x)-h(2x) \, \forall x$$



Now consider $P(x, 2x)$:



For $x \not =0$ we get $h(\frac{x}{2})=h(2x)-h(\frac{3x}{2})$. Since this holds for $x=0$ as well, we have $$h(\frac{x}{2})=h(2x)-h(\frac{3x}{2}) \, \forall x$$

Thus using the equation for $h(2x)$ above,


Finally consider $P(x, 3x)$:


Thus for $x \not =0$ we get $h(\frac{x}{2})-3h(\frac{3x}{2})=h(3x)-3h(x)-2h(2x)$. Since this holds for $x=0$ as well, we have

$$h(\frac{x}{2})-3h(\frac{3x}{2})=h(3x)-3h(x)-2h(2x) \, \forall x$$

Using the equation for $h(\frac{3x}{2})$ above,



Using the equation for $h(2x)$ above,




Replacing $x$ with $2x$ yields the desired equation $h(2x)=2h(x)$.

Step 5: Proving that $h(x)=xh(1)$.

Since $h(2x)=2h(x)$ we may now rewrite $P(x, y)$ as




Denote this as $Q(x, y)$.

We now perform a classic trick by using $Q(x, y), Q(x, 2y), Q(x+y, y)$ to write $h(x+2y)$ in two different ways, then comparing.

$Q(x+y, y)$ gives



Using $Q(x, y)$ gives



On the other hand, using $Q(x, 2y)$ gives


Since $h(2y)=2h(y)$, we have $$(x-2y)h(x+2y)=(x+4y)h(x)-(4x+4y)h(y)$$

Finally comparing gives

\begin{align} (x^2-xy)[(x+4y)h(x)-(4x+4y)h(y)] &=(x^2-xy)(x-2y)h(x+2y)\\ &=(x-2y)[(x^2+5xy+6y^2)h(x)-(4x^2+8xy)h(y)] \end{align}



Substitute $y=1$, so $h(x)=xh(1)$, as desired.

Step 6: Conclusion.

Let $h(1)=a$, so $h(x)=ax$. Then let $f(0)=b$, so $f(x)=h(x)+f(0)=ax+b$. Then $g(x)=c+\frac{1}{6}x(f(x)+f(\frac{x}{2})+f(0))=c+\frac{1}{6}x((ax+b)+(a\frac{x}{2}+b)+b)=\frac{1}{4}ax^2+\frac{1}{2}bx+c$. Finally, we may easily verify that these are solutions.

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If we additionally assume that $f$ is continuous, then by letting $y\to x$ we find that necessarily $g'(x)=\frac12 f(x)$. This immediately gives us a few solutions:

  • $g(x)=1, f(x)=0$
  • $g(x)=x, f(x)=2$
  • $g(x)=x^2, f(x)=4x$

and since the solutions form a vector space, any linear combination thereof. However, already with $g(x)=x^3, f(x)=6x$ the pattern breaks and the functional equation does not hold throughout!

Note that $g(x+h)-g(x-h) = \frac13 h(f(x+h)+f(x)+f(x-h)$ but also $g(x+h)-g(x-h) = g(x+h)-g(x)+g(x)-g(x-h)= \frac16 h(f(x+h)+f(x+h/2)+f(x))+\frac16h(f(x)+f(x-h/2)+f(x-h))$, hence $ f(x+h)+f(x-h)=f(x+h/2)+f(x-h/2)$ and by continuity at $x$, we conclude that $ f(x+h)+f(x-h)=2f(x)$ for all $h$. Then the right hand side of the f.e. is simply $\frac12(x-y)f(\frac{x+y}2)$. From this we have $g(x+h)-g(x-h)=hf(x)$. By subtracting a multiple of $x$ from $g$ and a corresponding multiple of $2$ from $f$, we may assume wlog. that $f(0)=0$, hence $g(x)=g(-x)$.

(to be continued)

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