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Determine the continuous function $f:\mathbb{R}\to\mathbb{R}$ with the following properties:

  1. $$\lim\limits_{x\rightarrow a} \frac{f(x)}{x-a}= e^{a},$$ where $a$ is a real valued constant;
  2. $$(x-a)(y-a)f(x+y)=(x+y-a)f(x)f(y),$$ for any $x, y \in \mathbb{R}$.

What I have tried

First of all, I have discovered the function $f_{1}=(x-a)e^{a}$, that obviously respects all criteria. I have a feeling that this is the only function.

Let's assume that $f_{2}$ is another function that checks all the criteria. Let $g:\mathbb{R}\to\mathbb{R}$, $$g(x) = \begin{cases}\dfrac{f_{1}(x)}{f_{2}(x)}, & x \neq a \\ 1, & \text{otherwise} \end{cases}$$.

Obviously, $g$ is also continuous. What I am trying to prove is that $g$ is constant. Does anyone have any ideas?

Math Enjoyer
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mathman
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  • A fixed variable is known as a constant. The word you want is continuous instead of continued. – Somos Mar 10 '23 at 21:27

1 Answers1

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You could rewrite the functional equation as $$\frac{f(x+y)}{x+y-a}=\frac{f(x)}{x-a}\cdot\frac{f(y)}{y-a}.$$ Define $g(x)=\frac{f(x)}{x-a}$. Then, the equation becomes $g(x+y)=g(x)g(y)$. Here, notice that $g$ would be continuous as well, being the product of continuous functions $f(x)$ and $\frac{1}{x-a}$ (you could argue that there could be problems at $x=a$, but they can be fixed). But then, it is a known result that the only continuous functions that satisfy this equation are $g(x)=e^{cx}$. But to satisfy the first condition as well, $c=1$ is needed. Then, $g(x)=e^x$, and $f(x)=(x-a)e^x$. The uniqueness is given by the functional equation which you can see more detailed in this post.

Math Enjoyer
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