Give an epsilon-delta proof to show that f ·g is continuous

Question

Let (M,d) be a metric space, let (V,||·||) be a real normed vector space, and let f : M →R and g : M → V be continuous. Give an epsilon-delta proof to show that f ·g is continuous, where · is scalar multiplication on V .

I am trying to define a delta for the norm of the the difference f·g to be less then epsilon. But here is where I stuck on. Could someone give a formal proof or provide any other possible way to show that? Many thanks?

Answer 1

Hint:

$$\begin{align}&|f(x)g(x) - f(x_0)g(x_0)| \\ =&\|f(x)g(x) - f(x_0) g(x) + f(x_0)g(x) - f(x_0)g(x_0)\|\\ \leq& |f(x)-f(x_0)|\cdot\|g(x)\| + |f(x_0)|\cdot\|g(x)-g(x_0)\|\\ \leq& M\cdot (|f(x)-f(x_0)| + \|g(x)-g(x_0)\|)\end{align}$$

for some value of $M$.

Now, picking $x_0$ close to $x$ will allow you to cause $|f(x)-f(x_0)|$ to be small, and $\|g(x)-g(x_0)\|$ to also be small.