$f,g:\mathbb{R}\to\mathbb{R}$ cont. on $\mathbb{R}$, $h\left( \left< x,y \right> \right) = \left< f(x),g(y) \right>$, $h$ is cont. on $\mathbb{R}^2$

Question

Let $f,g:\mathbb{R} \to \mathbb{R}$ be continuous on $\mathbb{R}$ and $h\left( \left< x,y \right> \right) = \left< f(x),g(y) \right>$, then $h$ is continuous on $\mathbb{R}^2$

Proof:

Let $\epsilon >0$ be given

$\forall x \in \mathbb{R}$

$\exists \ \delta_1>0$ s.t. for $a_1 \in \mathbb{R}$ s.t. $|f(x)-f(a_1)| \lt \frac{\epsilon}{\sqrt{2}}$ if $|x-a_1| \lt \delta_1$

$\exists \ \delta_2>0$ s.t. for $a_2 \in \mathbb{R}$ s.t. $|g(x)-g(a_2)| \lt \frac{\epsilon}{\sqrt{2}}$ if $|x-a_2| \lt \delta_2$

Choose $\delta_3 = \sqrt{\delta_1^2 + \delta_2^2}$

Let $\left<a_1, a_2\right> \in \mathbb{R}^2$ be a fixed point and $\forall \left< x_1,x_2 \right> \in \mathbb{R}^2$

$\sqrt{(x_1-a_1)^2 + (x_2-a_2)^2} \lt \sqrt{\delta_1^2 + \delta_2^2}=\delta_3$

Now, $\sqrt{\left(f(x_1)-f(a_1)\right)^2 + \left(g(x_2)-g(a_2)\right)^2}$ $\lt \sqrt{\frac{\epsilon^2}{2} + \frac{\epsilon^2}{2}} = \epsilon$

Since the point $\left< a_1,a_2 \right>$ was arbitrary, $h$ is continuous on $\mathbb{R}^2 \ \ \ \Box$

Answer 1

Your proof looks very good. Here is an alternate approach. Let, $\{(x_n,y_n)\}_{n\in\Bbb N}$ be a sequence in $\Bbb R^2$ converging to $(a,b)\in\Bbb R^2$, we show that $h(\langle x_n,y_n\rangle)\to h(\langle a,b\rangle)$.

Since $x_n\to a$ and $y_n\to b$ in $\Bbb R$ and $f,g$ are continuous in $\Bbb R$ so, $f(x_n)\to f(a)$ and $g(y_n)\to g(b)$. Also inner-product is continuous operator so $\langle f(x_n),g(y_n)\rangle \to \langle f(a),g(b)\rangle$ and hence $h(\langle x_n,y_n\rangle)\to h(\langle a,b\rangle)$. This shows that $h$ is continuous on $\Bbb R^2$.

Answer 2

It looks fine, but I do think you're getting a little confused about the definition of a continuous function.

If $(X,d_X)$ and $(Y, d_Y)$ are metric spaces, we say $f : A \subset X \to Y$ is continuous at $a \in A$ if for all $\epsilon > 0$ there exists $\delta > 0$ such that for all $x \in A$ if $d_X(x,a) < \delta$, then $d_Y(f(x), f(a)) < \epsilon$.

With this in mind, I would have structured your proof as follows:

Fix $\epsilon > 0$ and $(a_1, a_2) \in \mathbb{R}^2$.

Take $\delta_1 > 0$ such that for all $x \in \mathbb{R}$, $|x - a_1| < \delta_1$ implies $|f(x) - f(a_1)| < \frac{\epsilon}{\sqrt{2}}$.

Take $\delta_2 > 0$ and $\delta_2 > 0$ such that for all $x \in \mathbb{R}$, $|x - a_2| < \delta_2$ implies $|g(x) - g(a_2)| < \frac{\epsilon}{\sqrt{2}}$.

Let $\delta_3 = \min (\delta_1, \delta_2 ) > 0$. Fix $(x_1, x_2) \in \mathbb{R}^2$. Suppose $$ ||(x_1, x_2) - (a_1, a_2)|| = \sqrt{(x_1 - a_1)^2 + (x_2 + a_2)^2} < \delta_3 $$ Then $|x_1 - a_1| < \delta_3 \leq \delta_1$ and $|x_2 - a_2| < \delta_3 \leq \delta_2$. Then $|f(x) - f(a_1) | < \frac{\epsilon}{\sqrt{2}}$ and $|g(x) - g(a_2) | < \frac{\epsilon}{\sqrt{2}}$. Therefore $$ || h(x) - (a_1 , a_2) || = \sqrt{(f(x) - f(a_1))^2 + (g(x) - g(a_2))^2} < \sqrt{\frac{\epsilon^2}{2} + \frac{\epsilon^2}{2}} = \epsilon $$ Since $\epsilon > 0$ was arbitrary, we have shown that $h$ is continuous at $(a_1, a_2)$. Since $(a_1, a_2) \in \mathbb{R}^2$ was arbitrary, we have shown that $h$ is continuous on $\mathbb{R}^2$.

$\epsilon - \delta$ proofs are tricky because we first see why a proof will work by working backwards. But when the time comes to write up the proof, we must work forward.