Continuity of Max Function in R^2

[ttzhou]

This is my first post on PF, I've been a "Google lurker" for ages though, love the quality of the help provided here. I've done a search and found similar questions for when f, g are uniformly continuous and max(f,g) is discussed, but this question is purely for (x,y) in R^2. So hopefully, I haven't tread too heavily on my first post.

I tried typing LaTeX but it's been a nightmare, hopefully I'll work out the kinks in the future.

Homework Statement

Given h : R² ---> R, h(x,y) = max(x,y), show that h is continuous at any point (x₀,y₀) in its domain.

Homework Equations

There is an equation that relates max(x,y) to a function purely of the two variables, x and y. However, for the assignment, as well as my own curiosity, I wonder how one would go about doing this using the pure epsilon delta method for proving continuity, avoiding any characterization of continuity through sequences, etc.

The Attempt at a Solution

I've broken it down into cases.

1. If x₀ = y₀, then simply choose ||(x,y) - (x₀,y₀)|| < $$\delta = \epsilon$$.

Then, we note that, WLOG (we can interchange x₀ and y₀)

|max(x,y) - x₀| <= ||(x,y) - (x₀,x₀)|| = ||(x,y) - (x₀,y₀)|| < $$\delta = \epsilon$$

and so it is approximated by arbitrary epsilon whenever the delta condition is satisfied.


2. If x₀ does not equal y₀, this is where I get stumped.

I suppose WLOG that x₀ < y₀. (Again, I can switch around for the other case.)

Then |max(x,y) - max(x₀,y₀)| = |max(x,y) - y₀|

(Denote M = max(x,y) for convenience)

I cannot for the life of me figure out the proper algebra to get to a nice form where I can apply the appropriate deltas. There are too many messy attempts that I've made to list all here.

I tried |M - y₀| < |M - y| + |y - y₀| by triangle inequality, but that keeps leaving a mismatched x and y₀ or y and x₀, and I can't see how finding lower/upper bounds would work here either.

My professor has stated that I am on the right track with the cases, and gave a hint about using epsilon delta to play around with the inequality x₀ < y₀, but I can't figure it out.

I sketched the function mentally, it looks kind of like the sharp corner of the eaves of a house; I can see it geometrically, but am unable to find the right algebraic relation.

ANY help or hints are greatly appreciated! Thank you all in advance.

 

[tiny-tim]

welcome to pf!

hi ttzhou ! welcome to pf!

(have a delta: δ and an epsilon: ε and try using the X² and X₂ icons just above the Reply box )

Given h : R^2 ---> R, h(x,y) = max(x,y), show that h is continuous at any point (x0,y0) in its domain.
…
2. If x0 does not equal y0, this is where I get stumped …

yes 1 is ok (though a little messy)

for 2, just make sure your ε is less than |x₀ - y₀|

 

[ttzhou]

hi ttzhou ! welcome to pf!

yes 1 is ok (though a little messy)

Messy as in the proof itself or just the way I typed it? I tried typing $$\delta$$ and $$\epsilon$$, but it kept showing up as ||(x,y) - (x0,y0)|| [sic], just in LaTeX font.

Thanks for the tips, I'll take a crack at it later today.

EDIT: I'm having confusion as to what you mean by making $$\epsilon$$ < |x₀ - y₀|... Isn't $$\epsilon$$ reliant on $$\delta$$ and the point in question?

 

[tiny-tim]

Messy as in the proof itself … ?

Yes …

i] you don't choose (x,y), you choose δ (for any ε)

ii] then you should say for any ||(x,y) - (x₀,x₀)|| < δ ("for any" is the opposite of choosing, isn't it? )

iii] and you should have stuck to (x₀,x₀) (not (x₀,y₀)) throughout that proof

 

[ttzhou]

Yes …

i] you don't choose (x,y), you choose δ (for any ε)

This is my fault, I meant to say given ε > 0, choose δ = ε so that...

ii] then you should say for any ||(x,y) - (x₀,x₀)|| < δ ("for any" is the opposite of choosing, isn't it? )

whenever this condition is satisfied, the resulting approximation using ε follows.

iii] and you should have stuck to (x₀,x₀) (not (x₀,y₀)) throughout that proof

So I should have only changed it at the last line, since x₀ = y₀?


Also, still having trouble with the hint... How do I make ε less than something; isn't the approximation of ε dependent on what δ I choose?

 

[tiny-tim]

Also, still having trouble with the hint... How do I make ε less than something; isn't the approximation of ε dependent on what δ I choose?

yes, but you can choose ε to depend on δ and to be less than |x₀ - y₀|

 

[ttzhou]

yes, but you can choose ε to depend on δ and to be less than |x₀ - y₀|

We are given an arbitrary ε, and we must pick a δ such that the function is within ε of the image of (x₀,y₀)... Maybe it's the Tuesday afternoon haze, but I am just stumped as to how I can ensure ε to be less than anything, when we are only considering arbitrary values.

 

[tiny-tim]

sorry, i got my δ and my ε mixed up …

i meant make sure your δ is less than |x₀ - y₀| …

you can choose δ to depend on ε and to be less than |x₀ - y₀|

 

[ttzhou]

I may have arrived at a solution, but it seems unelegant:

Again, assume that x₀ < y₀ => 0 < y₀ - x₀

I choose δ = min { y₀ - x₀, (ε/4) }

Now, if ||(x,y) - (x₀, y₀)|| < δ, it must be the case that:

|x - x₀| < δ and |y - y₀| < δ

---

This is where I find it sort of inelegant, as I essentially prove that:

x₀ - y₀ < x - x₀ < y₀ - x₀ (by absolute value)

2x₀ - y₀ < x < y₀

2x₀ - 2y₀ < x - y₀ < 0

2(x₀ - y₀) < x - y₀ < 2(y₀ - x₀)

--> |x - y₀| < |2(y₀ - x₀)|

----

From here, I note that, denoting M = max(x,y):

|M - y₀|

<= |x - y₀| + |y - y₀|

<= |2(y₀ - x₀)| + |y - y₀|

< 2(ε/4) + (ε/4)

< ε

#


Does this look right? I just have my reservations about the inequality proof, I suppose I am just being an aesthetic-lover there though.

 

[tiny-tim]

I may have arrived at a solution, but it seems unelegant:

Again, assume that x₀ < y₀ => 0 < y₀ - x₀

I choose δ = min { y₀ - x₀, (ε/4) }

yeees … i think you meant "inelegant"!

the whole point of choosing that δ is so that you can start by saying that max(x,y) = y …

carry from there ​

 

[ttzhou]

I think what you mean is that pick a disk small enough so that any co-ordinate in that disk has the property that it lies in the region (y > x)...? But how does picking $$\delta$$ to be less than |x₀ - y₀| achieve this?

 

[tiny-tim]

I think what you mean is that pick a disk small enough so that any co-ordinate in that disk has the property that it lies in the region (y > x)...? But how does picking $$\delta$$ to be less than |x₀ - y₀| achieve this?

on second thought, it needs to be < |x₀ - y₀|/√2 …

draw (x₀,y₀), and also (y₀,y₀) on the line x = y …

then a circle of radius δ around (x₀,y₀) stays in the y>x region

 

[ttzhou]

Sigh... the funny thing is, I originally thought of a geometric approach, but never carried it through... It would have saved me so much time.

Yea, I was able to pinpoint that exact length using some Grade 7 geometry

Thanks a lot tiny-tim, really appreciate it!

P.S... how does one develop the ability to spot elegant solutions? I always have this tendency to bull my way through problems; they work, but I miss easier ways.

 

[tiny-tim]

Sigh... the funny thing is, I originally thought of a geometric approach, but never carried it through... It would have saved me so much time.

yup!

P.S... how does one develop the ability to spot elegant solutions? I always have this tendency to bull my way through problems; they work, but I miss easier ways.

always draw a diagram first … these proofs are all about points and circles anyway