How to find eigenfuctions of shifts operator and dilation operators

Question

I am trying to find eigenfunction of shift operators and dilation operators. Shift operators $S_t$ maps function $f(x)$ to $f(x+t)$. I am trying to find continuous function $f$ such that $(S_tf)(x) = \lambda_t f(x)$ for all $t\in\mathbb{R}$, i.e. $f(x+t)=\lambda_tf(x)$ for all $t$. I know the answer is $ce^{ax}$ for some $c,a$. It's easy to verify it, but how to prove it? In other words, how to prove that other continuous function can not be the eigenvector function?

I've shown that $f(0)$ is not 0 and I think we could also prove $f(x)=a^x$ for some a.

Similarly, I try to find eigenfunction of dilation operator which is defined as $(D_rf)(x)=f(rx)$ for $r>0$. I think they are similar, is there any way to solve them?

Answer 1

Shortly speaking, the continuous eigenfunction of all shifts operator has form $f(x)=ce^{ax}$ for some $c,a$ (complex is possible). The continuous eigenfunction of all dilation operator has form $f(x)=\alpha x^\beta$. Remark: the continuous condition is necessary here.

The proof is given in previous answer. I just want to simply check that they are correct.

$$(S_tf)(x) = f(x+t) = ce^{a(x+t)} = ce^{at}e^{ax} = \lambda_tf(x),$$ where $\lambda_t = ce^{at}$ is the corresponding eigenvalue. Thus $f(x)=ce^{ax}$ is eigenfunction of shifts operators.

$$(D_rf)(x) = f(rx) = \alpha (rx)^{\beta} = r^{\beta}\alpha x^{\beta} = \mu_rf(x),$$ where $\mu_r = r^\beta$ is the corresponding eigenvalue. Thus $f(x)=\alpha x^\beta$ is eigenfunction of dilation operators.

Answer 2

Suppose that $f: \mathbb{R} \to \mathbb{C}$ is a continuous eigenfunction for shift operators as you say, i.e. for every $t \in \mathbb{R}$ there is a $\lambda_t \in \mathbb{C}$ such that $(S_t f)(x) = f(x + t) = \lambda_t f(x)$ for all $x$. Also assume that $f(0) \neq 0$, as otherwise it's fairly quick to derive that $f \equiv 0$. Then the mapping $t \mapsto \lambda_t$ must be a continuous homomorphism from the reals under addition to the (nonzero) complex numbers under multiplication.

The homomorphism part (setting aside continuity) comes from noting that for any $s, t \in \mathbb{R}$, $$ \lambda_{s+t} f(0) = (S_{s+t} f) (0) = (S_s S_t f) (0) = \lambda_s \lambda_t f(0). $$ (Technically we should also show that no $\lambda_t$ can be equal to $0$, but I'll leave that as an exercise.) Being continuous comes from the continuity of $f$ together with the fact that $f(t) = \lambda_t f(0)$, i.e. $\lambda_t = f(t) / f(0)$.

It now follows (see here) that there is a complex number $z$ such that $\lambda_t = e^{zt}$. Therefore $f$ has the form $f(t) = f(0) e^{zt}$.

---

The question for dilations is fairly similar, though with a couple additional considerations. Here we begin by assuming that at least one of $f(1)$ and $f(-1)$ is nonzero, as otherwise we would obtain $f \equiv 0$. Using the notation $(D_r f)(x) = \mu_r f(x)$, $r > 0$, we can derive that $\mu_{rs} = \mu_r \mu_s$; and $f(r) = \mu_r f(1)$ and $f(-r) = \mu_r f(-1)$ for $r > 0$. So now the map $r \mapsto \mu_r$ is a continuous homomorphism from the positive reals under multiplication to the nonzero complex numbers under multiplication. But further, by the continuity of $f$ at $0$ we can conclude that the limit $\lim_{r \to 0} \mu_r$ must exist and be finite; and moreover that $f(1) = f(-1)$.

Via a logarithmic change of variable, we can convert to a continuous homomorphism from the reals under addition to the nonzero complex numbers under multiplication. The new homomorphism is given by $t \mapsto \mu_{e^t}$. As with the first part above for shift operators, we obtain a $z \in \mathbb{C}$ so that $\mu_{e^t} = e^{zt}$. Using the fact that $\lim_{r \to 0} \mu_r = \lim_{t \to -\infty} \mu_{e^t}$ exists and is finite, we conclude that $z$ must be a nonnegative real number. So now, denoting $e^z$ by $a$ (so that $a \geq 1$), we have $f(r) = f(1) a^{\log |r|}$ for $r \neq 0$, and $f(0)$ is either $f(1)$ (if $a = 1$) or $0$ (if $a > 1$).