For what continuous $f$ we can write $f(x)+a\cdot f(x+\alpha)$ as $b\cdot f(x+\beta)$?

Question

After reading about harmonic addition theorem, I am interested in:

Find all continuous $f:\mathbb R\mapsto\mathbb R$ that admit identities of the form $$f(x)+a\cdot f(x+\alpha)=b\cdot f(x+\beta)$$ for all real $a,\alpha,x$, where $b,\beta$ only depend on $a,\alpha$.

Obviously, we only have to determine $f$ up to a scaling constant and argument shift, because if $F(x)$ is such function, then so is $AF(x+B)$.

As a result of harmonic addition theorem, $\sin x$ is such function.

Also, the identity function $\text{Id}$ is also such function, with $b=a+1$, $\beta=\frac{a\alpha}{a+1}$.

Any other examples?

Can all such functions be found?

Answer 1

Interesting question. So when you found that $\sin x$ is such a function, and you mentioned argument shift, means that $\cos x$ is also a solution. Moreover, you can also have $\sin Cx$ and so on. This lead me to think about Fourier transforms. So let's multiply the equation by $e^{ikx}$ and integrate over $x$ from $-\infty$ to $\infty$ $$\int_{-\infty}^\infty f(x)e^{ikx}dx=F(k)\\\int_{-\infty}^\infty f(x+\alpha)e^{ikx}dx=\int_{-\infty}^\infty f(x)e^{ik(x-\alpha)}dx=e^{-ik\alpha}F(k)$$ Then we get $$F(k)+ae^{-ik\alpha}F(k)=be^{-ik\beta}F(k)$$ So it means that either $F(k)=0$ or $$1+ae^{-ik\alpha}=be^{-ik\beta}$$

Therefore, all the functions $f$ are the form $$f(x)=\sum_{k_j}c_{k_j}e^{-ik_jx}$$ where $k_j$ are the solutions of $1+ae^{-ik\alpha}=be^{-ik\beta}$.

It is easy to see that for $f(x)=\sin x$ we need to keep only $k=1$. Therefore $$1+a(\cos\alpha-i\sin\alpha)=b(\cos\beta-i\sin\beta)$$ We can write this as $$1+a\cos\alpha=b\cos\beta\\a\sin\alpha=b\sin\beta$$ If you plug these into the original equation, you get an identity.

Also note that for any constant function you get $k=0$, which translate into $1+a=b$

Edits: One can choose any $k_0$ and find $b$ and $\beta$ that solve the equation: $$1+a\cos(k_0\alpha)=b\cos(k_0\beta)\\a\sin(k_0\alpha)=b\sin(k_0\beta)$$ You get $b$ by squaring the equations and adding them together: $$b^2=1+a^2+2a\cos(k_0\alpha)$$ If $b\ne 0$ then by dividing the two equations you get $\tan\beta$.

With these $b$ and $\beta$, assume that also $k_1$ is a solution. By writing the equation for both and subtracting, you get $$a[\cos(k_0\alpha)-\cos(k_1\alpha)]=b[\cos(k_0\beta)-\cos(k_1\beta)]\\a[\sin(k_0\alpha)-\sin(k_1\alpha)]=b[\sin(k_0\beta)-\sin(k_1\beta)]$$ With some manipulation you might be able to get the possible values for $k_1$.

In conclusion, you choose a $k_0$, find out the corresponding $b$ and $\beta$, then find out any other $k_j$, and then finally find $f(x)$.

Answer 2

$$f(x):=e^{px}$$ is a solution when

$$1+a\,e^{p\alpha}=b\,e^{p\beta}.$$

This equation has an infinity of solutions, and $p$ complex is possible.

And by linearity,

$$\sum_{i=1}^n f_i\,e^{p_ix}$$ is also a solution.

---

Note that you can assume $a=1$ WLOG, because with $f(x):=e^{-(x\log a)/\alpha}g(x)$, the equation takes the form

$$e^{-(x\log a)/\alpha}g(x)+ae^{-(x\log a)/\alpha-(x\log a)/\alpha}g(x+\alpha)=be^{-(x\log a)/\alpha-(\beta\log a)/\alpha}g(x+\beta)$$

or $$g(x)+g(x+\alpha)=b\,a^{-\beta/\alpha}g(x+\beta)=b'g(x+\beta).$$

You can also assume $\alpha=1$ WLOG because by rescaling the variable ($\alpha t:=x$ and $h(t):=g(x)$),

$$h(t)+h(t+1)=b'h(t+\beta').$$