(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let [itex]a,b \in \mathbb{R}[/itex] and we define [itex]I = [a,b][/itex] ([] means closed set). Let [itex]\mathcal{C}_{\mathbb{K}}(I)[/itex] be the space of all continuous functions [itex]I \to \mathbb{K}[/itex] with the norm [itex]f \mapsto ||f||_I = \displaystyle \sup _{x \in I} |f(x)|[/itex]. Let [itex]U[/itex] be the set of all continuous functions [itex]I \to \mathbb{K}[/itex] so that [itex]\displaystyle |\int _a ^b f(x)dx| < 1[/itex]. Show that [itex]U[/itex] is an open subset of [itex]\mathcal{C}_{\mathbb{K}}(I)[/itex].

2. Relevant equations

If [itex]f: X \to Y[/itex] is a continuous function and [itex]B \subset Y[/itex] is open then the preimage [itex]f^{-1}[/itex] is open.

For [itex]B \subset Y[/itex] the set [itex]f^{-1}= \{x \in X : f(x) \in B\}[/itex] is called the preimage of B.

[itex]\mathbb{K} = \mathbb{R}[/itex] or [itex]\mathbb{C}[/itex]

3. The attempt at a solution

The function [itex]\mathcal{C} _{\mathbb{K}} (I) \to \mathbb{K}, f \mapsto \displaystyle \int _a ^b f(x)dx[/itex] is continuous because

[itex] \displaystyle |\int _a ^b f(x)dx| \leq \int _a ^b |f(x)|dx \leq \int _a ^b ||f||_I dx = ||f||_I \int _a ^b dx = (b-a)||f||_I[/itex]

I want to show that preimage of U is open then U is open using the two definitions I listed above. I'm just curious whether that is a legit method to solve this problem. I'm also wondering what the purpose of the norm is in this problem. If anyone can answer these questions I will be grateful.

Thanks.

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# Metric Spaces - Open subset

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