# Metrics on continuous functions question

1. Oct 2, 2012

### linda300

Hey guys,

I have been working on the following question:

http://imageshack.us/a/img407/4890/81345604.jpg [Broken]

For part a

f and g are continuous on I

=> there exists e > 0 and t_0 s.t.

0<|{f(t) - g(t)} - {f(t_0) - g(t_0)}| < e

using |a-b| >= |a| - |b|,

|{f(t) - g(t)} - {f(t_0) - g(t_0)}| >= |{f(t) - g(t)}| - |{f(t_0) - g(t_0)}|

and

|{f(t) - g(t)}| - |{f(t_0) - g(t_0)}| >= |{f(t) - g(t)}| - |f(t_0)| + |g(t_0)|

but i'm not sure where to go from here, I have tried a lot of different steps but I end up going in circles or getting back to the triangle inequality haha.

Does anyone have any ideas on how to do this? Is there some simple trick that I am just missing?

For part b,

Proving σ>=τ is pretty straight forward, just applying the Schwarz inequality but I'm having trouble showing that ρ>σ

How can you relate that sup to the integral expression σ?

Last edited by a moderator: May 6, 2017
2. Oct 2, 2012

### micromass

Staff Emeritus
For part (a): can you show that if $F:[0,1]\rightarrow \mathbb{R}$ is continuous and if $F(t_0)>0$ then there $F(t)\geq \frac{1}{2} F(t_0)$ for all t in some neighborhood of $t_0$.

Use continuity.

For (b): apply that $|f(t)-g(t)|\leq \rho(f,g)$ and use that inequality in the integral.

3. Oct 3, 2012

### linda300

So the continuity of F(t) gives you that (t_0 in [0,1])

For every ε>0 there exists a δ>0 such that |F(t) - F(t_0)|<ε for 0<|t-t_0|<δ

|F(t) - F(t_0)|≥|F(t)|-|F(t_0)|≥0

|F(t)|≥|F(t_0)|

I havn't been able to figure out how the 1/2 comes into it,

Is |F(t)|-|F(t_0)|≥0 not correct?

Is it just a matter of saying, then for t in [t_0 - d, t_0 + d] for some d it is possible that
|F(t)|-(1/2)|F(t_0)|≥0

4. Oct 3, 2012

### micromass

Staff Emeritus
It's not correct. Why should $|F(t)|-|F(t_0)|\geq 0$ hold? I agree that $|F(t)-F(t_0)|\geq |F(t)|-|F(t_0)|$ and that $|F(t)-F(t_0)|\geq 0$. But I don't see how that follows.

5. Oct 3, 2012

### linda300

Ah right

I'm really have trouble getting past this step

|F(t) - F(t_0)|≥|F(t)|-|F(t_0)|

Today I have been trying things like this,

|F(t)|-|F(t_0)|≥|F(t)|-2|F(t_0)|

|F(t)|-2|F(t_0)|≥(1/2)|F(t)|-2|F(t_0)|

|F(t)|-|F(t_0)|≥(1/2)|F(t)|-2|F(t_0)|

(1/2)|F(t)|-|F(t_0)|≥-2|F(t_0)|

|F(t)|≥-2|F(t_0)|

But keep getting unhelpful things like ^ haha

Is the next step something like what I did above?

6. Oct 3, 2012

### micromass

Staff Emeritus
You need to use continuity of F. Because F is continuous you know that: for each $\varepsilon>0$, there exists an interval I such that $t_0\in I$ and such that if $t\in I$, then $|F(t)-F(t_0)|<\varepsilon$.

Now, choose a special value for $\varepsilon$.

7. Oct 3, 2012

### linda300

I tried picking different ε before but I couldn't see how having

|F(t)| - |F(t_0)|≤|F(t) - F(t_0)|<ε

Could get to F(t)≥(1/2)F(t_0)

By saying F(t_0)>0 were you hinting that ε=F(t_0)?

8. Oct 3, 2012

### micromass

Staff Emeritus

$$|F(t_0)|-|F(t)|\leq |F(t_0)-F(t)|<\varepsilon$$
[/QUOTE]

Could get to F(t)≥(1/2)F(t_0)

By saying F(t_0)>0 were you hinting that ε=F(t_0)?[/QUOTE]

Something like that. What if you try $\varepsilon=|F(t_0)|/2$ ?

9. Oct 3, 2012

### linda300

Ohhhhhhhh

|F(t_0)| - |F(t)|<F(t_0)/2

Which is only true if

|F(t)|≥(1/2)|F(t_0)| !

Thanks heaps!