If $f(x.y)=f(x).f(y)$ for all $x,y$ and $f(x)$ is continuous at $x=1$,then show that $f(x)$ is continuous for all x except at $x=0$.Given $f(1)\neq 0$

Question

If $f(x.y)=f(x).f(y)$ for all $x,y$ and $f(x)$ is continuous at $x=1$,then show that $f(x)$ is continuous for all x except at $x=0$.Given $f(1)\neq 0$

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In the functional equation $f(x.y)=f(x).f(y)$,put $x=1,y=1$
$f(1)=f(1).f(1)\Rightarrow f(1)=1$ because $f(1)\neq 0$
As $f(x)$ is continuous at $x=1$,so $\lim_{x\to 1}=f(1)=1$
Now take any number $x_0\neq 0$
Put $x=x_0$ in the equation $f(x.y)=f(x).f(y)$
$\lim_{x\to1}f(x_0.x)=f(x_0).f(x)=f(x_0)\lim_{x\to1}f(x)=f(x_0).1=f(x_0)$
So i proved $\lim_{x\to1}f(x_0.x)=f(x_0)$

Therefore $f(x)$ is continuous at all non zero numbers.

But i do not know how to prove that $f(x)$ is not continuous at $x=0.$Please help me.Thanks.

Answer 1

Of course $f$ can be continuous at $x = 0$, as for example $f(x) = x^\alpha$ for $\alpha \ge 0$. (If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t shows that if $f$ is continuous at $x=0$ then it is necessarily of this form for $x \ge 0$.)

What probably is meant is

1. Show that for all $x \ne 0$, $f$ is continuous at $x$.
2. Show that $f$ is not necessarily continuous at $x=0$.

You have already done the first part (your proof works only for $x_0 \ne 0$). For the second part, one counter-example is sufficient: $$ f(x) = \frac 1x \text{ for } x \ne 0 \, , \quad f(0) = 0 $$ satisfies $f(xy) =f(x)f(y)$ and is continuous at $x=1$. But $f$ is not continuous at $x=0$.

Answer 2

Note first that $f(1)=f(1\cdot1)=f(1)\cdot f(1)=f(1)^2$. Since we asssume that $f(1)\neq 0$, dividing by $f(1)$ yields $f(1)=1$. Now $f(x)\cdot f(x^{-1})=f(x\cdot x^{-1})=f(1)=1$ from which it follows that $f(x^{-1})=f(x)^{-1}$ and hence $f(x/y)=f(x)/f(y)$.

To show that $f$ is continuous at $x \in \mathbf R \setminus \{0\}$, let $(x_n)$ be a sequence converging to $x$. We need to show that $f(x_n)\to f(x)$. Now since $x\neq 0$, we can consider the sequence $(y_n)$ defined by $y_n:=x_n/x$. Then $(y_n)\to 1$ and since $f$ is supposed to be continuous at 1, $f(y_n)\to f(1)=1$. Now $f(x_n)=f(y_n\cdot x)=f(y_n) \cdot f(x)\to 1\cdot f(x) = f(x)$, as was to be shown.

This exercise is a particular example of the following general fact: A homomorphism of topological groups is continuous iff it is continuous at one point.