## anonymous one year ago help here

1. anonymous

@Michele_Laino

2. anonymous

3. anonymous

4. Michele_Laino

here we have to keep in mind that: 1) the addition is a continuous function on R, namely R x R ---> R 2) the opposite value function is continuous in R, namely R--->R 3) the absolute value function is continuous in R, namely R---> R+ of course all those properties can be proved

5. Michele_Laino

so, using the theorem which states that the composition of two continuous functions is again a continuous function and the fact that the previous functions, namely max(x,y) and min(x,y) are continuous functions, then we get your thesis

6. Michele_Laino

in other words those three statements above allow us to say that max(x,y) and min(x,y) are continuous functions

7. ganeshie8

can w use epsilon-delta definition to prove the statements ?

8. Michele_Laino

I don't know, since f and g are generic functions

9. Michele_Laino

yes! we can prove those last three statements using epsilon-delta

10. Michele_Laino

nevertheless , since we can view this function: f(x)+g(x) as the composition of the functions: x---> (f(x),g(x)) and the addition function, then we can apply the theorem of composition of continuous function, in order to say that the function: f(x)+g(x) is a continuous function when also f and are continuous

11. ganeshie8

correct me if im wrong michele Since $$f$$ and $$g$$ are continuous at $$a$$, we have : $\forall \varepsilon > 0\ \exists \delta_f > 0\ \text{s.t. } 0 < |x -a| < \delta_f \implies |f(x) - f(a)| < \varepsilon$ $\forall \varepsilon > 0\ \exists \delta_g > 0\ \text{s.t. } 0 < |x -a| < \delta_g \implies |g(x) - g(a)| < \varepsilon$ let $$\delta_h=\min{\{\delta_f,~\delta_g\}}$$ and $$\delta_k=\max{\{\delta_f,~\delta_g\}}$$

12. Michele_Laino

ok!

13. ganeshie8

we want to show below : $$|x-a|\lt \delta_h \implies |h(x)-h(a)| \lt \epsilon$$ $$|x-a|\lt \delta_k \implies |k(x)-h(a)| \lt \epsilon$$

14. Michele_Laino

ok!

15. Michele_Laino

yes! I think that we can prove the statement using the epsilon-delta, if we consider these 2 cases, namely when f greter or equal to g, and when f less or equal to g. In those cases, we have: h(x)= f(x) and k(x) = g(x), and vice versa for second case, and by hypothesis both f and g are continuous

16. Michele_Laino

greater*

17. anonymous

is the proving still going on?

18. ganeshie8

I think the same $$\delta$$ works for both $$h(x)$$ and $$k(x)$$. Here is the complete proof : Since $$f$$ and $$g$$ are continuous at $$a$$, we have : $\forall \varepsilon > 0\ \exists \delta_f > 0\ \text{s.t. } 0 < |x -a| < \delta_f \implies |f(x) - f(a)| < \varepsilon ~~(\color{red}{\star})$ $\forall \varepsilon > 0\ \exists \delta_g > 0\ \text{s.t. } 0 < |x -a| < \delta_g \implies |g(x) - g(a)| < \varepsilon ~~(\color{red}{\diamond})$ Let $$\delta_h=\delta_k=\min{\{\delta_f,~\delta_g\}}$$ and suppose that $$f(a)\ge g(a)$$, then : $$|x-a|\lt \delta_h \implies$$ \begin{align}|h(x)-h(a)| &=|\max\{f(x),g(x)\}-\max\{f(a),g(a)\}|\\~\\ &=|\max\{f(x),g(x)\}-f(a)\}|\end{align} Case1 : $$f(x)\ge g(x)$$ $$|\max\{f(x),g(x)\}-f(a)\}|=|f(x)-f(a)|\lt \epsilon$$ by $$(\color{red}{\star})$$ Case2 : $$g(x)\ge f(x)$$ $$|\max\{f(x),g(x)\}-f(a)\}|=|g(x)-f(a)| \lt |g(x)-g(a)|\lt \epsilon$$ by $$(\color{red}{\diamond})$$ similarly we can prove the other statement by considering two cases

19. Michele_Laino

referring to my proof we have finished. So reassuming, in order to prove your statement, we need of the subsequent properties: 1) the addition is a continuous function on R, namely R x R ---> R 2) the opposite value function is continuous in R, namely R--->R 3) the absolute value function is continuous in R, namely R---> R+ furthermore, using the theorem of composition of continuous functions, and the fact that this function: x--->(f(x),g(x)) is continuous, then we get your thesis

20. ganeshie8

Please double check.. it might contain errors as im really bad with real analysis...

21. Michele_Laino

that's right! @ganeshie8

22. Michele_Laino

I was thinking to your method :)

23. ganeshie8

xD

24. Michele_Laino

now you have two different solutions @GIL.ojei

25. anonymous

thank you sirs but is that all from you @ganeshie8

26. ganeshie8

I think what I had there proves that $$h(x)$$ is continuous at $$a$$. I'm hoping you can mimic the same easily for proving continuity of $$k(x)$$ too...

27. anonymous

I WILL BUT IT WOULD BE MORE EASIER FOR ME IF YOU COMPLETE IT IN OTHER NOT TO HAVE ERRORS, COS DON'T KNOW HOW TO USE THE SYMBOLS IN THIS SITE

28. anonymous

@ganeshie8

29. ganeshie8

(... continued) $$|x-a|\lt \delta_k \implies$$ \begin{align}|k(x)-k(a)| &=|\min\{f(x),g(x)\}-\min\{f(a),g(a)\}|\\~\\ &=|\min\{f(x),g(x)\}-g(a)\}|\end{align} Case1 : $$f(x)\ge g(x)$$ $$|\min\{f(x),g(x)\}-g(a)\}|=|g(x)-g(a)|\lt \epsilon$$ by $$(\color{red}{\star})$$ Case2 : $$g(x)\ge f(x)$$ $$|\min\{f(x),g(x)\}-g(a)\}|=|f(x)-g(a)| \lt |g(x)-g(a)|\lt \epsilon$$ by $$(\color{red}{\diamond})$$ $$\blacksquare$$

30. anonymous

@ganeshie8

31. anonymous