If $f$ is a continuous real function defined on a metric space $X$, then $\alpha f$ is also continuous, where $\alpha$ is any real number.

Question

Let $\epsilon > 0$ be given. By the continuity of $f$ we know $\exists \delta: d(x,x') < \delta \implies |f(x)-f(x')| < \epsilon / |\alpha|$. Then the continuity of $\alpha f$ at $x'$ can be seen by

$d(x,x') < \delta \implies |(\alpha f)(x)-(\alpha f)(x')| = |\alpha(f(x)-f(x'))| = |\alpha| |f(x)-f(x')| < |\alpha| * \epsilon/|\alpha| = \epsilon$.

Therefore $\alpha f$ is continuous at $x'$, and because $x'$ is arbitrary $\alpha f$ is continuous everywhere.

Note: I tagged vector-spaces because my textbooks says that the set $L$ of all real valued functions defined on a metric space $X$ is a vector space. This problem is to prove that $L$ is closed under scalar multiplication.

Thank you for your help.

Answer 1

It is correct, if $\alpha\neq0$. If $\alpha=0$, then you get the null function, which is clearly continuous.

It remains to be proved that if $f$ and $g$ are continuous, then $f+g$ is continuous too.