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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 $d(f, f_0)$?

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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.

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  • $\begingroup$ 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

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