Proof that $ f: X \times \mathbb{R} \to X: (x,\lambda) \mapsto \lambda x$ is continuous

Question

Let $(X, \|\cdot\|)$ be a normed space, and $\|\cdot\|_M : X \times \mathbb{R} \to \mathbb{R}^+: (x,y) \mapsto \max(\|x\|,|y|)$.

Proof that $ f: X \times \mathbb{R} \to X: (x,\lambda) \mapsto \lambda x$ is continuous by using the $\epsilon$ - $\delta$ defintion of continuity.

This is what I've got so far:

Take $(a,b) \in X \times \mathbb{R}$, $\forall \epsilon >0$, $\exists \delta >0$, $\forall (x,y) \in X \times \mathbb{R}$:

$\|(x,y)-(a,b)\|_M=||(x-a,y-b)\|_M=\max(\|x-a\|,|y-b|)<\delta$

$\|f(x,y)-f(a,b)\|=\|yx-ab\|=\|y(x-a)+a(y-b)\| \le |y|\|x-a\| + \|a\||y-b|$

I don't know what to do now, because I don't have any information about $|y|$ and $\|a\|$...

Answer 1

As you've already shown, we can bound $$ |\lambda x - \alpha y| < |\lambda| \delta + |y| \delta < |\lambda| \delta + (\delta + |x|) \delta < 2M \delta + \delta^2 $$ where $M = \max(|x|, |\lambda|)$. Now choose $\delta < M$. Then $$ |\lambda x - \alpha y| < 3M \delta, $$ in which case we can guarantee that $|f(x,\lambda) - f(y, \alpha)| < \epsilon$ by choosing any $\delta$ satisfying $\delta < \epsilon /(3M)$ and $\delta < M$.