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Conjecture and prove a statement about a differentiable function satisfying f(x+a) = bf(x)

by RoRi

Let f(x) be a function which is differentiable everywhere and which satisfies

    \[ f(x+a) = bf(x) \]

for some positive constants a and b. What can you conclude about such a function f?

(Note: I’m not entirely sure what Apostol wants here since the instruction “what can you conclude” is pretty vague. He does give an “answer” in the back of the book, so I verify that it does have the properties indicated, but I don’t know how you would arrive at that expression just from the question statement. I’ll mark this question as incompletely and hopefully come up with something better in the future.)

Since f(x) satisfies the functional equation f(x+a) = bf(x) we can write

    \[ f(x) = b^{\frac{x}{a}} g(x) \]

which implies

    \[ g(x) = \left( \frac{1}{b} \right)^{\frac{x}{a}} f(x). \]

Then computing

    \begin{align*} g(x+a) &= \left( \frac{1}{b} \right)^{\frac{x+a}{a}} f(x+a) \\ &= \left(\frac{1}{b}\right)^{\frac{x}{a}} \cdot \frac{1}{b} \cdot bf(x) \\ &= \left( \frac{1}{b} \right)^{\frac{x}{a}} f(x) \\ &= g(x). \end{align*}

Thus, g(x) is indeed periodic with period a and so

    \begin{align*} f(x+a) &= b^{\frac{x+a}{a}} g(x+a) \\ &= b\cdot b^{\frac{x}{a}} g(x) \\ &= b f(x). \end{align*}

So, this definition of f(x) in terms of the periodic function g(x) indeed satisfies the functional equation.

2 comments

  1. Luca says:

    I think you can derive the form from the back of the book. Assuming f satisfies the functional equation, we may define
    g(x)=b^(-x/a)f(x), and we then find that
    g(x+a)=b^((-x/a)-1)f(x+a)=b^((-x/a)-1)*bf(x)=b^(-x/a)f(x),
    and conclude that g is periodic with period a.

  2. Mohammad Azad says:

    This is circular reasoning, you used the fact that f(x+a)=bf(x) to prove that g is periodic and then used that to prove that f(x+a)=bf(x) !!!

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