# Thread: Real Analysis and Metric Spaces question

1. ## Real Analysis and Metric Spaces question

Let (X,dx) and (Y,dy) be two metric spaces and f: X-->Y be a mapping satisfying the Lipshitz condition

dy(f(x),f(y))<(or equal to) Ldx(x,y), L does not equal zero, for all x,y elements of X.

is this uniform continuous and if so how would does anyone know what proof would be used well to prove it?

i'll check it this site very frequently so any help will be appreciated. thnx.

2. ## Re: Real Analysis and Metric Spaces question

Just use the $\varepsilon$ / $\delta$ definition of the limit. If you take $\varepsilon>0$, you can find a $\delta$ which works (it can be expressed with $\varepsilon$ and $L$).

3. ## Re: Real Analysis and Metric Spaces question

thank you. the epsilon delta definition is where you put in a modulus the function - the limit? i'll look it up now coz i haven't felt that comfortable with the epsilon delta definition in the past really.

4. ## Re: Real Analysis and Metric Spaces question

you couldnt show me how could you im pretty bad at epsilon delta definitions. i just can't really fidn anywhere that givesa good consie explanation... thnx.

5. ## Re: Real Analysis and Metric Spaces question

btw why do you guys like helping so much. is it because you're like phd's and it helps you revise what you know or keep you fresh on it? or does having a good rank on here look good on your cv or what?

6. ## Re: Real Analysis and Metric Spaces question

Originally Posted by maxgunn555
Let (X,dx) and (Y,dy) be two metric spaces and f: X-->Y be a mapping satisfying the Lipshitz condition
dy(f(x),f(y))<(or equal to) Ldx(x,y), L does not equal zero, for all x,y elements of X. is this uniform continuous and if so how would does anyone know what proof would be used well to prove it?
Because $L>0$, if $c>0$ then so is $\tfrac{c}{L}$.
In the definition let $\delta=\tfrac{c}{L}$.

Now it just falls out.

7. ## Re: Real Analysis and Metric Spaces question

so 0< $\mid\ x\ -\ c\mid\ <\ c/L\rightarrow \mid\ f(x)-L\mid <\ \epsilon$

8. ## Re: Real Analysis and Metric Spaces question

otherwise is see limf(x) =f(c) as x approaches c but i don't see how that related to the epsilon delta definition. ie what i'd substitute in.