Dear Romeo Meštrović

The same argument for the one variable case holds.

You can extend the results as follows

f(x,y)=(±1)ⁿ f(x+nb,y),n∈Z

f(x,y)=(±1)^{k} f(x,y+kd),k∈Z

f(x,y)=(±1)^{n+k} f(x+nb,y+kd),n,k∈Z

One should be careful about all possible variations

where a,c belong { 1, -1}.

Examples: a = c =1, double periodic (2b,2d).

f(x,y)=sin((π/b)x)sin((π/d)y)

f(x,y)=sin((π/b)x)+sin((π/d)y)

f(x,y)= 2((x/(2b))-⌊(1/2)+(x/(2b))⌋)+ 2((y/(2d))-⌊(1/2)+(y/(2d))⌋)

f(x,y)= 4((x/(2b))-⌊(1/2)+(x/(2b))⌋)((y/(2d))-⌊(1/2)+(y/(2d))⌋)

f(x, y) = - af(x + b, y) = - cf(x, y +d) for all x, y ϵ R.

Check the first one:

f(x+b,y)=sin((π/b)(x+b))sin((π/d)y)= - f(x,y)

f(x,y+d) = sin((π/b)x)sin((π/d)(y+d))= - f(x,y)

i.e f(x,y) = - f(x+b,y) = - f(x,y+d).

There are an infinite number of solutions for the proposed 2- variable functional equation.

Best regards

Any solution will be bounded. Suppose f not identical 0

Let M be the max of | f | and T the period of the function.

If | a | > 1 then in the interval say [b, b + T] there is y such that | f (y) | = M

Then | f (y - b) | = | a | |f ( y ) | > M, contradiction

Therefore in this case there is no solution other than the constant 0.

If |a| < 1 you can get a similar contradiction.

So you can have solutions only for a = - 1 and a = 1

For a = - 1 any periodic continuous function such that b is a period is a solution.

If a = 1 there are other type of solutions. If b = pi the trigonometric functions are solutions

For general b you can scale the trig functions to get solutions

How about a sine function, a=1, b=pi ?

sin(x+pi)=sin(x) cos( pi) + cos(x) sin(pi) = -sin(x)

And, to finish the above proof, note that sin (πx/b) is a solution in the case a=1, i.e., continuous with period 2b.

Thank you for your help, it was fast and rather easy.

Dear my colleagues

I think the following set

S={ f in C such that f(x)+af(x+a)=0}

is closed space of C. So, it is possible that S include an infinite dimension space. It must verify by one.

Thanks

You can consider the function f(x)=exp(ix) with a=1 and b=(2k+1)π, k∈Z.

I think each solution of main problem could approximate by Fourier series.

A suggestion to solve this functional equation is the following fact

• $$f(x)=(-1)^n a^n f(x+nb),\;\;\forall n\in\Z_+$$

Mr. Mesrizadeh:

It is indeed an infinite dimensional vector space for all nonzero a & b

One example how to construct an infinite dimensional linearly independent set when a and b are positive.

Let p be a prime number. We construct a function f_p on [0, b]

At 0, b/p, 2b/p, ...b the function is 0. In between two consecutive zeros the function is a tent on the positive side. For example it goes linearly from (0,0) to

(b/2p, 1) and then linearly from (b/2p, 1) to (b/p, 0) and goes the same way on all other intervals.

Then on [b, 2b] we define it as tents on the negative side such that f(x) + a f(x + b) = 0 for all x in [0, b]. Notice that there is an unique way to do it as long as f is defined on [0, b]. We continue in the same way with positive tents on [2b, 3b] and so on and then we go on the [-b, 0] and so on.

All these functions are linearly independent because each of them has a zero (b/p) that neither one of the others have.

Dear George Stoica

On the solution of the functional equation f(x) + af(x+b) = 0.

We have f(x) = -af(x+b)

then f(x+b) = -af(x+2b), and f(x+2b) = - af(x+3b), etc.

In general, f(x+nb)= -af(x+(n+1)b)

by direct substitution, we obtain

f(x)= -a(-af(x+2b))= a²f(x+2b)=a²(-af(x+3b))=-a³f(x+3b)

and then f(x)= (-a)ⁿ f(x+nb), for any n ∈ N (*)

(*) This relation was observed by Daniel Alfonso Santiesteban .

Similarly, the formula works for n ∈ Z and not only for Z+.(check)

Therefore f(x) = (-a)ⁿ f(x+nb), n ∈ Z.

If |a| 1, f(x) → ∞ (*)

(*) The above results were observed by Gabriel T. Prajitura

using different approach.

So, the only possible solution is when |a| = 1

For a = 1, f(x) = (-1) f(x+b) = f(x +2b)

any periodic function of period 2b is a solution.

Examples: f(x)= sin((π/b)x)

f(x)= 2(x/2b- ⌊1/2+ x/2b⌋ )

where ⌊* ⌋ is the floor function.

but if a = -1, we have f(x) = f(x+b),

then any function of period b is a solution.

Examples:

f(x)= sin((π/b)x)

f(x)= 2(x/b- ⌊1/2+ x/b⌋ )

Conclusion: The functional equation f(x) + af(x+b) = 0 has infinite number of solutions iff a = ± 1.

Best regards

for any x in R exist n in Z, suth that y=x-nb in [0,b).

Let \phi(y) any function (y\in [0,b) and a not equal 0 (if a=0 only solution f(x)=0). Then f(x)=(-1/a)^n\phi(y) is solution.

f(x) is continiuosly only if \fhi(y) continiously and lim_{y\to b}\phi(y)=-\phi(0)/a.

f(x) periodic function only if (-1/a)^m=1, then period is m*b. If f real function, then m=1 (a=-1) or m=2 (a=1). If f(x) complex function, m in N (any natural).

\phi(y) continiuosly, \phi(0)=\phi(b), then f(x) is periodic (a=-1, f(x+b)=f(x)) with period b.

\phi(y) continiously, \phi(0)=-\phi(b), then f(x) is periodic with period 2b

f(x+b)=-f(x)\to f(x+2b)=-f(x+b)=f(x).

I give all functions, suth that f(x)+af(x+b)=0.

Minimal period can be less than b. For example a=1

f(x)=siin(\pi xm/b)

Say b > 0. Let C_0[0, b] be the space of all continuous functions f from [0, b] to R such that f(0) = f(b).

The equation f (x) + a f (x + b) = 0 allows as shown before an unique extension to R. Call this extension F_f.

The function

H : C_0 [0, b] to C_ab (R) given by H (f) = F_f is an invertible linear operator.

Here C_ab (R) is the set of all functions from g R to R satisfying g (x) + a g (x + b) = 0.

Therefore C_ab (R) is an infinite dimensional subspace (I should say linear manifold though) of C (R).

I don't know what topology to consider on C (R) but the supremum norm does not work and definitely does not work on C_ab (R) because if |a| not = 1 except for the function 0 all functions in this space are unbounded.

However, we can define a norm on C_ab (R) by || F_f || = || f ||, so by transporting the norm on C_0 [0, b] where the sup norm is OK.

With this norm C_ab (R) is a Banach space.

One last thing. C_0 [0, b] is in fact a Banach algebra. C_ab (R) is not, unless a = - 1.

That is because F_f F_g not equal F_fg so only the linear structure transfers but not the multiplicative one.

In the previous answer I missed something.

If should be f (b) = - (1/a) f (0) instead of f(b) = f (0).

R. R. Aidagulov

Have you shown any addition to the previous answers?

With pervious answer of my colleagues, extremely way can exist to make the solution of main problem. One may use Fourier series with unknown coefficients that it satisfies in f(x)+af(x+b)=0. We must look infinite unknown coefficients with infinite number equations, indeed multiplicity of solutions are shown by one.

• If a0 and f satisfies the proposed functional equation then h(x)= ax f(bx) is continuous and 2-periodic (h(x+2)=-h(x+1)=h(x)), in particular, it is bounded. Now, if a is not equal to +1 and f is d-periodic with f(x0) different from zero, then (h((x0+k d)/b))kεZ cannot be bounded which is absurd. So, either f is identically zero, or a=1.
• Now when a=1, let g(x)=(-1)[x] f(bx) where [x] is the integer part of x then clearly g is 1- periodic continuous on R\Z and satisfies g(0)=g(0+)—g(0–)=—g(1–).
• Conversely, if g:[0,1]—>R is continuous and satisfies g(0)+g(1)=0 then f(x)=(-1)[x/b]g({x/b}) where {x} is the fractional part of x is a solution of the proposed functional equation satisfying the required conditions.

A two-variable analogue of the proposed problem can be given as follows. Given real numbers, a ≠ 0, b, c ≠ 0 and d, find all continuous on R^2 real-valued functions f satisfying the equations

f(x, y) = - af(x + b, y) = - cf(x, y +d) for all x, y ϵ R.

Dear Romeo Meštrović

The same argument for the one variable case holds.

You can extend the results as follows

f(x,y)=(±1)ⁿ f(x+nb,y),n∈Z

f(x,y)=(±1)^{k} f(x,y+kd),k∈Z

f(x,y)=(±1)^{n+k} f(x+nb,y+kd),n,k∈Z

One should be careful about all possible variations

where a,c belong { 1, -1}.

Examples: a = c =1, double periodic (2b,2d).

f(x,y)=sin((π/b)x)sin((π/d)y)

f(x,y)=sin((π/b)x)+sin((π/d)y)

f(x,y)= 2((x/(2b))-⌊(1/2)+(x/(2b))⌋)+ 2((y/(2d))-⌊(1/2)+(y/(2d))⌋)

f(x,y)= 4((x/(2b))-⌊(1/2)+(x/(2b))⌋)((y/(2d))-⌊(1/2)+(y/(2d))⌋)

f(x, y) = - af(x + b, y) = - cf(x, y +d) for all x, y ϵ R.

Check the first one:

f(x+b,y)=sin((π/b)(x+b))sin((π/d)y)= - f(x,y)

f(x,y+d) = sin((π/b)x)sin((π/d)(y+d))= - f(x,y)

i.e f(x,y) = - f(x+b,y) = - f(x,y+d).

There are an infinite number of solutions for the proposed 2- variable functional equation.

Best regards

Dear Issam Kaddoura,

Of course. And also in the case of functions of several ( n ) real variables (with related n functional equations).

Best regards

For f(0) non-zero and t positive number,

f(x)=f(0)exp((2.\pi.i.x)/t) is a solution, with a=exp(-2.\pi.i.b)/t) and f(x+t)=f(x).

This particular case may give some idea for generalization.

There is no a 1-periodic function with

f(x)=f(x+\sqrt(2)) since all m+m\sqrt(2) are dense in R

Based on intuition alone sounds like a line with positive slope with superimposed

periodic function; that way you could get an identity on the same equivalent pair of points of the curve.