Suppose $a$ and $b$ are distinct real numbers and $f$ is a continuous real function such that $\frac{f(x)}{x^2}$ goes to 0 when $x$ goes to infinity or minus infinity. Suppose that$ f(x+a)+f(x+b)=\frac{f(2x)}{2}$. How show that $f$ is periodic?

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    $\begingroup$ Hmm... can you name any nontrivial function with this property? $\endgroup$ Commented Dec 28, 2014 at 19:00
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    $\begingroup$ Unfortunately, I do not know $\endgroup$
    – piteer
    Commented Dec 28, 2014 at 19:05
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    $\begingroup$ My guess (quite uneducated in functional equations) that somehow we would be able to prove that the only such function will be constantly zero. $\endgroup$ Commented Dec 28, 2014 at 19:14
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    $\begingroup$ Assuming $f$ is periodic, it must be constant, since the minimal period of the left hand side must be twice the minimal period of the right hand side. And since $f(2a) + 2f(a+b) = f(2b) + 2f(a+b) = 0$, this constant must be zero. $\endgroup$ Commented Dec 28, 2014 at 19:24
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    $\begingroup$ if you put x=a , x=b ypu will have $$(1) f(2a)+2f(a+b)=0\\(2) f(2b)+2f(a+b)=0\\(1)-(2): f(2a)-f(2b)=0\\f(2a)=(2b)\\$$ a and b are distinct real numbers so I think period = |2a-2b| $\endgroup$
    – Khosrotash
    Commented Dec 28, 2014 at 19:41

2 Answers 2


There are not so many possibilities for equation transformations, so we go straight forward. For the determinancy assume that $a<b$.

Let $x$ be an arbitrary real number. Then



$$f(2x+a)=2\left(f\left(x+a+\frac a2\right)+ f\left(x+b+\frac a2\right)\right),$$

$$f(2x+b)=2\left(f\left(x+a+\frac b2\right)+ f\left(x+b+\frac b2\right)\right).$$

So $$f\left(x+a+\frac a2\right)+ f\left(x+b+\frac a2\right)+f\left(x+a+\frac b2\right)+ f\left(x+b+\frac b2\right)=\frac{f(4x)}{4}.$$

Similarly, by induction we can show that for each non-negative integer $n$

$$S_n(x)= \sum_{i=0}^{2^n-1} f\left(x+2a+\frac{ib-(i+1)a}{2^{n-1}}\right)=\frac{f(2^nx)}{2^n}.$$

Indeed, above this equality is already proved for $n\le 2$. Assume that the equality is proved for each real $x$ and a non-negative integer $n$. Then

$$\frac{f(2^{n+1}x)}{2^{n+1}}=\frac{1}{2^n}\left(f(2^nx+a)+ f(2^nx+b)\right)=$$ $$S_n\left(x+\frac{a}{2^n}\right)+ S_n\left(x+\frac{b}{2^n}\right)=$$ $$\sum_{i=0}^{2^n-1} f\left(x+\frac{a}{2^n}+2a+\frac{ib-(i+1)a}{2^{n-1}}\right)+ f\left(x+\frac{b}{2^n}+2b+\frac{ib-(i+1)a}{2^{n-1}}\right)=$$

$$\sum_{i=0}^{2^n-1} f\left(x+2a+\frac{2ib-(2i+1)a}{2^n}\right)+ f\left(x+2a+\frac{(2i+1)b-(2i+2)a}{2^n}\right)=$$ $$ \sum_{i=0}^{2^{n+1}-1} f\left(x+2a+\frac{ib-(i+1)a}{2^n}\right)=S_{n+1}(x).$$

Let $$\sigma_n(x)=\frac{b-a}{2^n} \sum_{i=0}^{2^n-1} f\left(x+2a+\frac{i(b-a)}{2^{n-1}}\right)$$ be an integral sum for the function $f$ at the segment $[x+2a, x+2b]$. Since the function $f$ is continuous on this segment, it is integrable on it, so for each real $x$ a sequence $\{\sigma_n(x)\}$ converges to an integral $\int^{x+2b}_{x+2a} f(t)dt$. Let $\tau_n(x)=\frac{b-a}{2^n}S_n(x)$. Since the function $f$ is continuous on the segment $I=[x+\min\{0,2a\}, x+\max\{0,2a\}+2(b-a)]$ , it is uniformly continuous on it. Then for each $\varepsilon>0$ there exists natural $N$ such that if $n>N$, $t,t’\in I$, and $|t-t’|<\frac{|a|}{2^{n-1}} $ then $|f(t)-f(t’)|<\varepsilon$. Then

$$|\tau_n(x)-\sigma_n(x)|=$$ $$\left|\frac{b-a}{2^n}\sum_{i=0}^{2^n-1} f\left(x+2a+\frac{ib-(i+1)a}{2^{n-1}}\right) - \frac{b-a}{2^n} \sum_{i=0}^{2^n-1} f\left(x+2a+\frac{i(b-a)}{2^{n-1}}\right)\right|=$$

$$\frac{b-a}{2^n}\left|\sum_{i=0}^{2^n-1} f\left(x+2a+\frac{ib-(i+1)a}{2^{n-1}}\right) - f\left(x+2a+\frac{i(b-a)}{2^{n-1}}\right)\right|< $$

$$\frac{b-a}{2^n}\sum_{i=0}^{2^n-1}\varepsilon=\frac{(b-a)}{2^n}\cdot 2^n\varepsilon=(b-a)\varepsilon.$$

Thus the sequence $\{\tau_n(x)\}$ converges to the integral $\int^{x+2b}_{x+2a} f(t)dt$ for each real $x$.

One the other hand


tends to zero when $n$ tends to infinity (the case $x\ne 0$ directly follows from the question condition, in the case $x=0$ it suffices to remark that $f(2^nx)=f(0)$ for all $n$).

Thus $\int^{x+2b}_{x+2a} f(t)dt=0$ for each real $x$. Differentiating both sides of this equality with respect to $x$ (this is allowed, by instance, by [Fich, 305, 12$^\circ$, p. 116]), we obtain $f(x+2b)-f(x+2a)=0$. Since $a\ne b$, the function $f$ is periodic. Since it is continuous, it is constant or has a minimal period. By Hans Engler’s comment the latter case is impossible and this constant is zero.

What’s a pity, so much efforts for zero!


[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, vol. II, 7-th edition, M.: Nauka, 1970 (in Russian).


We can substitute x=a-x and x=b-x, We get $$2(f(2a-x)+f(a-x+b)=f(2a-2x), 2(f(b-x+a)+f(2b-x)=f(2b-2x)$$ Here subtracting these two equations we get, $$2(f(2a-x)-2f(2b-x))=f(2a-2x)-f(2b-2x)$$ Putting x=0, we get, $$f(2a)=f(2b)$$ Also from the equation, $$f(2a-2x)-2f(2a-x)=f(2b-2x)-2f(2b-x)$$

Lets call the above equation Z, we could substitute x as x/2, and multiply by 2 on both sides.

We get, $$2(f(2a-x)-2f(2a-\frac{x}{2}))=2(f(2b-x)-2f(2b-\frac{x}{2}))$$

Adding this equation with Z we get, $$f(2a-2x)-2^2f(2a-\frac{x}{2})=f(2b-2x)-2^2f(2b-\frac{x}{2})$$

We could generalise this for n, $$f(2a-2x)-2^nf(2a-\frac{x}{2^{n-1}})=f(2b-2x)-2^nf(2b-\frac{x}{2^{n-1}})$$ When n tends to infinity, $$f(2a-2x)-2^nf(2a)=f(2b-2x)-2^nf(2b)$$ As $$f(2a)=f(2b),$$ $$f(2a-2x)=f(2b-2x)$$ Substituting x=a-b in this equation we get, $$f(2b)=f(4b-2a)$$

We can see that $$f(2a)=f(2b)=f(4b-2a)$$ From this we can see the period is $$2b-2a$$

Please do correct me if wrong.


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