Show that if $\pi_i \circ f$ is continuous for each i where $f$ is a function $f: \mathbb{R}^n \rightarrow{} \mathbb{R}^m$, $f$ is continuous

Question

Show that if $\pi_i \circ f$ is continuous for each i where $f$ is a function $f: \mathbb{R}^n \rightarrow{} \mathbb{R}^m$$\iff$ $f$ is continuous.

Where $\pi_1, ..., \pi_m: \mathbb{R}^m \rightarrow \mathbb{R}$ is defined by $\pi_i(\mathbf{x}) = x_i$ for $\mathbf{x} = (x_1,...,x_m) \in \mathbb{R}^m$.

So I show $\pi_i$ is continuous to show one direction. Want to show $||\bar{x} - \bar{a}|| < \delta \implies ||f(\bar{x}) - f(\bar{a})|| < \epsilon$. Fix $\epsilon > 0$. Take $\bar{z} \in \mathbb{R}^m, ||\bar{x} - \bar{z}|| < \epsilon \implies \sqrt{(x_1 - z_1)^2 + \cdots + (x_n - z_n)^2} < \epsilon \\ \implies (x_i - z_i)^2 < \epsilon^2 \implies |x_i - z_i| < \epsilon$.

So take $\delta = \epsilon$. $||f(\bar{x}) - f(\bar{z})|| < \epsilon \implies ||\pi_i(\bar{x}) - \pi_i(\bar{z})|| < \epsilon \implies |x_i - z_i| < \epsilon$. And so $\pi_i$ is continuous. Since $f$ is continuous and the composition of continuous functions is continuous, we have that $\pi_i \circ f$ is continuous. So one direction is shown.

Stuck on showing the other direction, that $\pi_i \circ f \implies f$ continuous. My current idea is to do a proof by contradiction assuming $f$ is discontinuous, that $\exists \epsilon > 0, \forall \delta > 0, \exists x \in S$ where $|x-a| < \delta$ but $|f(x) - f(a)| \geq \epsilon$. But getting stuck on the details.

Answer 1

It's all a consequence of the simple fact that for arbitrary ${\bf x}$, ${\bf y}\in{\mathbb R}^m$ one has $$\pi_i({\bf x})=x_i\qquad(1\leq i\leq m)$$ and then $$|y_i-x_i|\leq |{\bf y}-{\bf x}|\leq\sum_{k=1}^m |y_k-x_k|\ .$$ Here the second inequality is a consequence of the triangle inequality in ${\mathbb R}^m$. It follows that for arbitrary ${\bf a}$, ${\bf b}$ in the domain of ${\bf f}$ the following inequalities are valid: $$|f_i({\bf b})-f_i({\bf a})|\leq\bigl|{\bf f}({\bf b})-{\bf f}({\bf a})|\leq \sum_{k=1}^m |f_k({\bf b})-f_k({\bf a})|\qquad(1\leq i\leq m)\ .$$ This shows that $\bigl|{\bf f}({\bf b})-{\bf f}({\bf a})|$ is small, e.g., smaller than a prescribed $\epsilon>0$, iff all $|f_i({\bf b})-f_i({\bf a})|$ are sufficiently small.

Answer 2

To see that $\pi_i(x)$ is continuous let $\delta = \epsilon>0$. We have,

$$\|\pi_i(x)-\pi_i(y)\| = |x_i-y_i| = \sqrt{(x_i-y_i)^2}\leq \sqrt{\sum\limits_{i=1}^m (x_i-y_i)^2}=\|x-y\|<\delta=\epsilon$$

where the inequality is because $g(z)=\sqrt{z}$ is an increasing function.

You already know how this implies $\pi_i\circ f$ is continuous.

To see that $\pi_i\circ f$ continuous $\forall i\implies$ $f$ continuous, assume $\pi_i\circ f$ continuous $\forall i$ and let $\epsilon>0$. Then, $\forall i$, $\exists \delta_i>0$ such that

$$\|x-y\|<\delta_i\implies|f(x)_i-f(y)_i|<\frac{\epsilon}{\sqrt{m}}$$

Let $\delta = \min\limits_{i}\delta_i$. Then, $\forall y:\|x-y\|<\delta$, $$\|f(x)-f(y)\| = \sqrt{\sum\limits_{i=1}^m |f(x)_i - f(y)_i|^2}< \sqrt{\sum\limits_{i=1}^m\left(\frac{\epsilon}{\sqrt{m}}\right)^2} = \sqrt{m}\frac{\epsilon}{\sqrt{m}}=\epsilon$$

where the inequality is again because of monotonicity.