# Continuity of a projection mapping

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 $$d(f, f_0)$$?

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.
• 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 $$d(f, g) \geq \sum_{n = n_0}^{\infty} \frac{1/k}{2^n} = \frac{2^{n_0+1}}{k}$$ May 14 at 14:33