Question

help here

Answer 1

@Michele_Laino

Answer 2

please help with question number 5b

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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 \(\delta_k=\max{\{\delta_f,~\delta_g\}}\)

Answer 11

ok!

Answer 12

we want to show below :

\(|x-a|\lt \delta_h \implies |h(x)-h(a)| \lt \epsilon\)
\(|x-a|\lt \delta_k \implies |k(x)-h(a)| \lt \epsilon\)

Answer 13

ok!

Answer 14

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

Answer 15

greater*

Answer 16

is the proving still going on?

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

that's right! @ganeshie8

Answer 21

I was thinking to your method :)

Answer 22

xD

Answer 23

now you have two different solutions @GIL.ojei

Answer 24

thank you sirs but is that all from you @ganeshie8

Answer 25

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...

Answer 26

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

Answer 27

@ganeshie8

Answer 28

(... 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 \((\color{red}{\diamond})\)

\(\blacksquare\)

Answer 29

@ganeshie8