To clarify the given question, we have $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ linear projection. How can we prove that we can find $M \in \mathbb{R}$ such that \begin{align} ||f(u)|| \leq M ||u||, \ \forall u \in \mathbb{R}^{n}? \end{align} To clarify notation, we use $||.||$ for the Euclidean distance formula for $m$-dimensions. For example $||x|| = \sqrt{x_1^2 + \ldots + x_m^{2}}$. We must also prove that $f$ is continuous. That every linear projection is continuous is trivial I think?
To prove the inequality, shouldn't we have any sort of upper bound to tie it to?
Any help would be appreciated!
@Mark's answer gave me a hint, and I took the rest.
We can prove that \begin{align} |T(x)| \leq M \end{align} when $|x|=1$.
Taking the standard basis for $\mathbb{R}^m$ we may write \begin{align} x = a_1e_1 + \ldots + a_me_m \end{align} where $|a_i| \leq 1 \ \forall \ 1 \leq i \leq m$. So we have \begin{align} |T(x)| &= |T(a_1e_1+\ldots+a_me_m)|\\ &= |a_1 T(e_1) + \ldots + a_m T(e_m)|\\ &\leq |a_1| |T(e_1)| + \ldots + |a_m| |T(e_m)|\\ &\leq |T(e_1)| + \ldots+|T(e_m)| = M. \end{align}
So there exists $M>0$ such that if $h \neq 0$ we have \begin{align} \big| T \left( \dfrac{h}{|h|} \right) \big| \leq M\\ \implies |\dfrac{1}{|h|} T(h)| \leq M\\ \implies |T(h)| \leq M|h|. \end{align}
To prove that a linear transformation is continuous, it suffices to notice \begin{align} |T(h)| \leq M |h| \end{align} so if $\epsilon > 0$ is given and $h \in \mathbb{R}^n$ then whenever $0 < |x-h| < \dfrac{\epsilon}{M}$ we can have \begin{align} |T(x) - T(h)| = |T(x-h)| \leq M |x-h| < M \dfrac{\epsilon}{M} = \epsilon. \end{align} Therefore by definition the linear projection is continuous.