# Prove the function is continuous (topology)

1. Jun 21, 2013

### DotKite

1. The problem statement, all variables and given/known data
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. Relevant 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

2. Jun 21, 2013

### verty

Find a counterexample if you think it is untrue.