Is $f(x+y)=f(x)f(y)$ $\forall x,y \in \mathbb{R}$ continuous at $0 \implies $ continuos at $a$?(proof verification)

Question

This is a simple proof of Real Analysis in a single variable.

Given $f(x+y)=f(x)f(y)$ $\forall x,y \in \mathbb{R}$

Show that if $f $ is continuos at $x=0$ then $ f$ is continuos at$ x=a$.

Hint or suggestion: make a change of variable $u=x-a$.

Please verify my proof:

Let $lim_{x \rightarrow 0} f(x+y)=f(x)f(y)$ So $lim_{x \rightarrow 0} f(x+y)=f(0)f(y)$

Let $\epsilon>0$ exists such that $\delta>0$

We have: $0<|x-0|< \delta$ $\implies$
$$|f(x+y)-f(x)f(y)|<\epsilon$$

(by definition of continuity)

So, let $u=x-a \implies u\rightarrow x-a$

Such that $x\rightarrow 0$ and $u-a \rightarrow 0$ $\implies u\rightarrow a$

Then, we get the limit:

$$lim_{u \rightarrow a} f(u-a+y)=f(u-a)f(y)$$

Therefore $\epsilon>0$ exists when $\delta>0$ such that $0<|u-a|<\delta$ it implies $$|f(u-a+y)-f(u-a)f(y)|<\epsilon$$

$\therefore f $ is continuos at $x=0 \implies$ $f$ is continuous at $x=a_{\blacksquare}$ Is it correct?.

Suggestions will be welcome. Thanks in advance.

Answer 1

@Γιάννης Παπαβασιλείου already gave a nice answer, I add another interesting fact. It is known that the exponential $f(x) = a^x$ is the unique function with $f(1) = a$ that satisfies $f(x+y)=f(x)f(y)$. Since it is continuous you have your statement.

Answer 2

Let $x_0\in\mathbb{R}$. Then $\lim\limits_{x\to x_0}f(x)= \lim\limits_{u\to0}f(u+x_0)=\lim\limits_{u\to0}[f(u)f(x_0)]=\lim\limits_{u\to0}f(u)\lim\limits_{u\to0}f(x_0)=f(0)f(x_0)=f(0+x_0)=f(x_0)$ so $f$ is continous at $x=x_0\checkmark$