Proof that a non-zero function $ f: \mathbb{R} \to \mathbb{R} $ is continuous if $ f(x + y) = f(x) f(y) $ and $ f $ is continuous at one point. Let $ f: \mathbb{R} \to \mathbb{R} $ be a function such that $ f(x + y) = f(x) f(y) $ for all $ x,y \in \mathbb{R} $ and $ f $ is not identically zero. Show that if there exists a point at which $ f $ is continuous, then $ f $ is continuous on all of $ \mathbb{R} $.
I can easily see why $ f(x) \neq 0 $ for all $ x \in \mathbb{R} $, but that doesn’t seem to help. I also tried a proof by contradiction but seemed to get nowhere, so any help is greatly needed and would be immensely appreciated. Thanks.
 A: The idea is that, if f is continuous at a, you can prove it's continuous at 0, then at every x :
First, remark that f(0) = 1
Second, if a is continuous at a, then 
$$ \lim_{h\to 0} f(a+h) = f(a)$$
and
$$ \lim_{h\to 0} f(a+h) = f(a)\left( \lim_{h\to 0} f(h) \right)$$
This means that 
$$  \lim_{h\to 0} f(h) =1 = f(0)$$
Hence f is continuous at 0
But if f is continuous at 0, take $b \in \mathbb{R}$, and you have
$$ \lim_{h\to 0} f(b+h) = f(b)\left( \lim_{h\to 0} f(h) \right) = f(b)f(0) = f(b)$$
and f is continous at b

A proof with $\epsilon$ and without using the fact that $f(0)=1$
Let $\epsilon > 0$. As $f$ is continous at $a$, it exist $\delta >0$ such that for all $h \in] -\delta, \delta[$, we have $|f(a+h)-f(a)|<\epsilon$. Then
$$|f(b+h)-f(b)| = |f(a+h+(b-a))-f(b)| = |f(a+h)f(b-a) - f(b)| $$
Now suppose that $f(a+h)\geq f(a)$ and $f(b-a) \geq 0$ (it's the same for the other cases)
$$\leq |(f(a)+\epsilon)f(b-a) - f(b)| = |\epsilon f(b-a) + f(a)f(b-a) - f(b)| $$
$$= |\epsilon f(b-a) + f(b) - f(b)| = \epsilon |f(b-a)|$$
And f is continuous at b
A: This is a special case of a general theorem that says:

Let $ G $ and $ H $ be Hausdorff topological groups and $ \phi: G \to H $ a group homomorphism. If $ \phi $ is continuous at some point, then it is continuous everywhere.

The proof is simple. Suppose that $ \phi $ is continuous at $ g \in G $, i.e., $ \displaystyle \lim_{x \to g} \phi(x) = \phi(g) $, and let $ a \in G $ be another point. Then
\begin{align}
    \lim_{x \to a} \phi(x)
& = \lim_{x \to a} \phi(x) \phi(a^{-1} g g^{-1} a) \\
& = \lim_{x \to a} \phi(x a^{-1} g) \phi(g^{-1} a) \\
& = \left[ \lim_{x \to a} \phi(x a^{-1} g) \right] \phi(g^{-1} a) \quad
    (\text{By the continuity of group multiplication in $ H $.}) \\
& = \left[ \lim_{y \to g} \phi(y) \right] \phi(g^{-1} a) \quad
    (\text{By the continuity of group multiplication in $ G $.}) \\
& = \phi(g) \phi(g^{-1} a) \qquad (\text{By the continuity of $ \phi $ at $ g $.}) \\
& = \phi(a).
\end{align}
In our case, the topological groups are (as $ f $ is not identically $ 0 $)
$$
G = (\mathbb{R},+,0_{\mathbb{R}}) \quad \text{and} \quad
H = (\mathbb{R}_{> 0},\times,1_{\mathbb{R}}),
$$
and $ f $ is a group homomorphism from $ G $ to $ H $. As $ f $ is assumed to be continuous at a point, it must be continuous everywhere.
