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Prove the function is continuous (topology)

  • Thread starter DotKite
  • Start date
  • #1
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1. Homework Statement
Let X be the set of continuous functions ## f:\left [ a,b \right ] \rightarrow \mathbb{R} ##.
Let d*(f,g) = ## \int_{a}^{b}\left | f(t) - g(t) \right | dt ## for f,g in X. For each f in X set,

## I(f) = \int_{a}^{b}f(t)dt ##

Prove that the function ## I ##: (X,d*) → (X,d) is continuous

2. Homework Equations

The definition of d is given as d(## a_{i},b_{i} ##) = max ## \left \{ \left | a_{i} - b_{i} \right | \right \} ##

Here is the definition of continuity I am working with

Let (X,d) and (Y,d') be metric spaces, and let a be in X. A function f : X→Y is said to be continuous at the point a in X if ## \varepsilon > 0 ## and there exists a ## \delta > 0 ## such that

d'(f(x),f(a)) < ## \varepsilon ##

whenever x is in X and

d(x,a) < ## \delta ##


3. The Attempt at a Solution

Let ## \varepsilon > 0 ##, there exists a ## \delta > 0 ## such that if d*(f,g) < ## \delta ##

then

d(I(f), I(g)) = max {|I(f) - I(g)|} = max ## \left \{ \left | I(f)-I(g) \right | \right \} ## = max ## \left \{ \left | \int_{a}^{b} f(t)dt - \int_{a}^{b}g(t) dt \right | \right \} ## = max ## \left \{ \left | \int_{a}^{b} f(t) - g(t) dt \right | \right \} ##...


So I was thinking that i should let ## \delta = \varepsilon ## and somehow show

max ## \left \{ \left | \int_{a}^{b} f(t) - g(t) dt \right | \right \} ## < ## \delta ##

I could show that by showing max ## \left \{ \left | \int_{a}^{b} f(t) - g(t) dt \right | \right \} ## < ## \int_{a}^{b}\left | f(t)-g(t) \right |dt ## because d(f,g) < ## \delta ##

But I don't think that inequality is true, and this is where I am stuck
 

Answers and Replies

  • #2
verty
Homework Helper
2,157
198
Find a counterexample if you think it is untrue.
 

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