Let $C$ denote the space of continuous functions $f : [ 0, \infty ) \rightarrow \mathbb{R}^d$ equipped with the metric $$ d (f,g) = \sum_{n = 1}^{\infty} \left( 1 \wedge \sup_{x \in [0, n]} |f(x) - g(x)| \right)2^{-n}. $$

For $a > 0$, let $C'$ denote the space of continuous functions $f : [ 0, a ] \rightarrow \mathbb{R}^d$ equipped with the metric induced by the uniform norm $$ \lVert f \rVert_{\infty} = \sup_{ x \in [ 0, a ] } |f(x)|. $$

Is the mapping $$ R : C \rightarrow C', \quad f \mapsto R ( f ) = f \big\vert_{[0, a]} $$ continuous?

Let $\varepsilon > 0$ and $f_0 \in C$ be fixed. If $f \in C$ is such that $d(f, f_0)<\varepsilon$, is it possible to show that $$ \lVert R(f) - R(f_0) \rVert_{ \infty } = \sup_{ x \in [ 0, a ] } |f(x) - f_0(x)| $$ is less than some multiple of $d(f, f_0)$?


1 Answer 1


Let $a\le n_0.$ Assume $d(f,g)<{ 1\over k2^{n_0}},$ $k\ge 1.$ Then $|f(x)-g(x)|\le {1\over k}$ for $0\le x\le n_0.$ Indeed, if $$|f(x_0)-g(x_0)|>{1\over k},\quad {\rm for\ some}\ 0\le x_0\le n_0,$$ then $$d(f,g)\ge [1\wedge \max_{0\le x\le n_0}|f(x)-g(x)|]\, 2^{-n_0}>{1\over k2^{n_0}}$$ Hence $$\sup_{0\le x\le a}|f(x)-g(x)|\le \sup_{0\le x\le n_0}|f(x)-g(x)|\le {1\over k}$$ This gives continuity of the projection mapping.

  • $\begingroup$ Did you mean to write $d(f, g) < \frac{2^{n_0}}{k}$? Because if we assume that the condition $|f(x)-g(x)| \leq \frac{1}{k}$ for $0 \leq x \leq n_0$ is violated, one would get $$ d(f, g) \geq \sum_{n = n_0}^{\infty} \frac{1/k}{2^n} = \frac{2^{n_0+1}}{k} $$ $\endgroup$
    – Harry
    May 14 at 14:33
  • $\begingroup$ I meant what has been stated in my answer. I have extended the explanation to make it clearer. $\endgroup$ May 14 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.