MHB How can we show the existance of a ε?

[mathmari]

Hey!

Suppose that $h_1,h_2:[0,1]\rightarrow \mathbb{R}$ be continuous functions.

I want to show that there is a $\epsilon>0$ such that $|h_1(x)-h_2(x)|\geq \epsilon$ for each $x\in [0,1]$, given that $h_1(x)\neq h_2(x)$ for each $x\in [0,1]$. I have done the following:

We know that $h_1(x)\neq h_2(x)$ for each $x\in [0,1]$, that means that $f(x):=h_1(x)-h_2(x)$ is either $>0$ or $<0$ but never $=0$.

Therefore, we have that $|f(x)|>0, \forall x\in [0,1]$. That means that there is a $\epsilon>0$ such that $|f(x)|\geq \epsilon, \forall x\in [0,1]$. Is this correct? (Wondering)

 

[Euge]

Yes, it's correct. However, you may need to justify the last step. Since $|f|$ is continuous on a closed and bounded interval, it has a minimum at some point $a\in [0,1]$. Setting $\epsilon = |f(a)|$ does the job.

 

[mathmari]

Yes, it's correct. However, you may need to justify the last step. Since $|f|$ is continuous on a closed and bounded interval, it has a minimum at some point $a\in [0,1]$. Setting $\epsilon = |f(a)|$ does the job.

I see... Thank you very much! (Yes)