# find all function f,g that satisfy

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? meta.math.stackexchange.com/questions/5020/… –  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

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

$$\frac{1}{6}x(f(x)+f(\frac{x}{2})+f(0))-\frac{1}{6}y(f(y)+f(\frac{y}{2})+f(0))=\frac{1}{6}(x-y)(f(x)+f(\frac{x+y}{2})+f(y))$$

Some rearranging gives

$$(x-y)f(0)+xf(\frac{x}{2})-yf(\frac{y}{2})=xf(y)-yf(x)+(x-y)f(\frac{x+y}{2})$$

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

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

Then the equation becomes

$$xh(\frac{x}{2})-yh(\frac{y}{2})=xh(y)-yh(x)+(x-y)h(\frac{x+y}{2})$$

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

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

$P(x, -x)$ gives

$$xh(\frac{x}{2})+xh(-\frac{x}{2})=xh(-x)+xh(x)+2xh(0)$$

$$xh(\frac{x}{2})+xh(-\frac{x}{2})=xh(-x)+xh(x)$$

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

$$h(\frac{x}{2})+h(-\frac{x}{2})=h(-x)+h(x)$$

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

$$h(\frac{x}{2})+h(-\frac{x}{2})=h(-x)+h(x)$$

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

$$-xh(-\frac{x}{2})+yh(-\frac{y}{2})=-xh(-y)+yh(-x)+(y-x)h(-\frac{x+y}{2})$$

$$xh(-\frac{x}{2})-yh(-\frac{y}{2})=xh(-y)-yh(-x)+(x-y)h(-\frac{x+y}{2})$$

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

$$xk(\frac{x}{2})-yk(\frac{y}{2})=xk(y)-yk(x)+(x-y)k(\frac{x+y}{2})$$

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

$$xk(x)-yk(y)=xk(y)-yk(x)+(x-y)k(x+y)$$

$$(x-y)k(x+y)=(x+y)k(x)-(x+y)k(y)$$

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

$$-xk(3x)=3xk(x)-3xk(x)=0$$

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

$$xh(\frac{x}{2})+2xh(-x)=xh(-2x)+2xh(x)+3xh(-\frac{x}{2})$$

$$xh(\frac{x}{2})-2xh(x)=-xh(2x)+2xh(x)-3xh(\frac{x}{2})$$

$$4xh(\frac{x}{2})=4xh(x)-xh(2x)$$

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

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

Rearranging,

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

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

$$xh(\frac{x}{2})-2xh(x)=xh(2x)-2xh(x)-xh(\frac{3x}{2})$$

$$xh(\frac{x}{2})=xh(2x)-xh(\frac{3x}{2})$$

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,

$$h(\frac{3x}{2})=h(2x)-h(\frac{x}{2})=[4h(x)-4h(\frac{x}{2})]-h(\frac{x}{2})=4h(x)-5h(\frac{x}{2})$$

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

$$xh(\frac{x}{2})-3xh(\frac{3x}{2})=xh(3x)-3xh(x)-2xh(2x)$$

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,

$$h(\frac{x}{2})-3[4h(x)-5h(\frac{x}{2})]=[4h(2x)-5h(x)]-3h(x)-2h(2x)$$

$$16h(\frac{x}{2})=2h(2x)+4h(x)$$

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

$$16h(\frac{x}{2})=2(4h(x)-4h(\frac{x}{2}))+4h(x)$$

$$24h(\frac{x}{2})=12h(x)$$

$$h(x)=2h(\frac{x}{2})$$

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

$$\frac{1}{2}xh(x)-\frac{1}{2}yh(y)=xh(y)-yh(x)+\frac{1}{2}(x-y)h(x+y)$$

$$xh(x)-yh(y)=2xh(y)-2yh(x)+(x-y)h(x+y)$$

$$(x-y)h(x+y)=(x+2y)h(x)-(2x+y)h(y)$$

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

$$xh(x+2y)=(x+3y)h(x+y)-(2x+3y)h(y)$$

$$x(x-y)h(x+2y)=(x+3y)[(x-y)h(x+y)]-(2x+3y)(x-y)h(y)$$

Using $Q(x, y)$ gives

$$x(x-y)h(x+2y)=(x+3y)[(x+2y)h(x)-(2x+y)h(y)]-(2x+3y)(x-y)h(y)$$

$$(x^2-xy)h(x+2y)=(x^2+5xy+6y^2)h(x)-(4x^2+8xy)h(y)$$

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

$$(x-2y)h(x+2y)=(x+4y)h(x)-(2x+2y)h(2y)$$

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}

$$(x^3+3x^2y-4xy^2)h(x)-(4x^3-4xy^2)h(y)=(x^3+3x^2y-4xy^2-12y^3)h(x)-(4x^3-16xy^2)h(y)$$

$$12y^3h(x)=12xy^2h(y)$$

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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