Let $X$ be a Banach space and $A$ be a nonempty closed subsets of $X$.
Consider the following function: $(t,x)\mapsto tx$ from $[0,1]\times A$ to $X$. Show that it is continous function:
My effort:
Let $(a,x) $ and $(b,y)$ in $[0,1]\times A$, we have: $$ \|ax-by\|\leq \|ax-ay\|+\|ay-by\|=|a|\|x-y\|+|a-b|\|y\| $$ then, if $(a,x) $ tends towards $(b,y) $ we have: $$ \|ax-by\|\to 0 $$ Then this function is continous.
This is true.?
Let $(X,||\cdot||)$ a normed vector space (not necessarily a Banach space) and the map \begin{equation*} \begin{split} f : [0,1] \times A & \to X \\ (\alpha, x) & \mapsto \alpha x \end{split} \end{equation*} where $A\subset X$ is a non-empty closed subset of $X$. Then, \begin{equation*} \begin{split} ||f(\alpha,x)-f(\beta, y)|| & =||\alpha x-\beta y|| \\ & =||\alpha x-\alpha y+ \alpha y-\beta y|| \\ & \leq ||\alpha x-\alpha y||+|| \alpha y-\beta y|| \\ & = |\alpha|\cdot ||x-y||+||y||\cdot|\alpha-\beta| \end{split} \end{equation*} Since $A$ is closed, then for every $x\in A$ there exists a sequence $\{x_n\}\subset A$ that converges to $x$. Analogously, since $[0,1]$ is closed, then for every $\alpha \in A$ there exists a sequence $\{\alpha_n\}\subset [0,1]$ that converges to $\alpha$.
Hence, for every $\varepsilon>0$ we can pick $y\in A\backslash \{0\}$ and $\beta\in [0,1]$ such than $$||x-y||\leq \frac{\varepsilon}{2|\alpha|}, \quad |\alpha-\beta|\leq \frac{\varepsilon}{2||y||}$$ and the previous inequality is written as $$ ||f(\alpha,x)-f(\beta, y)|| \leq |\alpha|\cdot ||x-y||+||y||\cdot|\alpha-\beta|\leq |\alpha|\frac{\varepsilon}{2|\alpha|}+||y||\frac{\varepsilon}{2||y||} = \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$
Observe that this argument does not hold when $\alpha=0$. However, we can write for this case \begin{equation*} \begin{split} ||f(\alpha,x)-f(\beta, y)|| & =||\alpha x-\beta y|| \\ & =||\alpha x-\beta x+ \beta x-\beta y|| \\ & = |\beta|\cdot ||x-y||+||x||\cdot|\alpha-\beta| \end{split} \end{equation*} and use an analogous argumentation.