I'm trying to show that $h(x,t) = tx_0 + (1-t)x$ is continuous for $x \in \Bbb R ^n$ and $t \in \Bbb R$.
I can prove that if $f,g: X \rightarrow \Bbb R^n$ for a metric space $X$ are continuous then $f + g$ is continuous, but I'm having a hard time showing:
If $f : X \rightarrow \Bbb R^n$ and $g : Y \rightarrow \Bbb R$ are continuous, then $g*f$ is continuous.
I see that the continuity of $h$ follows from the two facts above but I'm having trouble making any progress.
Anyone have any ideas?
Best Answer
Hint: The proof is essentially the same as in Euclidean space. We want to show that $f\cdot g$ is continuous in the point $(x_1,y_1)\in X\times Y$. Now pick $(x_2,y_2)\in X \times Y$, then we compute $$ \vert f(x_1)g(y_1) - f(x_2)g(y_2) \vert = \vert f(x_1)(g(y_1)-g(y_2)) + (f(x_1)-f(x_2))g(y_2) \vert \leq \vert f(x_1) \vert \cdot \vert g(y_1) - g(y_2) \vert + \vert f(x_1) - f(x_2) \vert \cdot \vert g(y_2) \vert \leq \vert f(x_1) \vert \cdot \vert g(y_1) - g(y_2) \vert + \vert f(x_1) - f(x_2) \vert \cdot \vert g(y_2) - g(y_1) \vert + \vert f(x_1) - f(x_2) \vert \cdot \vert g(y_1) \vert.$$ Now use the continuity of $f$ and $g$. Can you finish from here?
Added: Let $\varepsilon >0$, by the continuity of $f,g$ there exists $\delta_f, \delta_g>0$ such that $d_X(x_1, x_2)<\delta_f$ and $d_Y(y_1,y_2)< \delta_g$ implies $\vert f(x_1) -f(x_2) \vert < \min\{ \frac{\varepsilon}{2(1+ \vert f(x_1)\vert + \vert g(y_1) \vert)}, 1\}$ and $\vert g(y_1) - g(y_2) \vert < \min\{ \frac{\varepsilon}{2(1+ \vert f(x_1)\vert + \vert g(y_1) \vert)}, 1\}$. Using the inequality above, we get $$ \vert f(x_1)g(y_1) - f(x_2)g(y_2) \vert \leq \vert f(x_1) \vert \cdot \vert g(y_1) - g(y_2) \vert + \vert f(x_1) - f(x_2) \vert \cdot \vert g(y_2) - g(y_1) \vert + \vert f(x_1) - f(x_2) \vert \cdot \vert g(y_1) \vert \leq (\vert f(x_1) \vert + \vert g(y_1) \vert) \min\{ \frac{\varepsilon}{2(1+ \vert f(x_1)\vert + \vert g(y_1) \vert)}, 1\} + \left(\min\{ \frac{\varepsilon}{2(1+ \vert f(x_1)\vert + \vert g(y_1) \vert)}, 1\}\right)^2 < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$ Hence, we for $d_{X\times Y}((x_1,y_1);(x_2,y_2)) < \min\{ \delta_f, \delta_g\}$ we have (as $d_{X\times Y}((x_1,y_1);(x_2,y_2))= d_X(x_1,x_2) + d_Y(y_1,y_2)$) that $$ \vert f(x_1)g(y_1) - f(x_2)g(y_2) \vert < \varepsilon.$$ Thus, the function $$ f\cdot g : (X\times Y, d_{X\times Y}) \rightarrow (\mathbb{R}, \vert \cdot \vert), \ (x,y) \mapsto f(x) \cdot g(y) $$ is continuous at the point $(x_1,y_1)$. However, this point was arbitrary, hence, $f\cdot g$ is continuous.