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Simple but frustrating confusion on definition of bounded function

by HJ Farnsworth
Tags: bounded, confusion, definition, frustrating, function, simple
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HJ Farnsworth
#1
Feb9-11, 12:28 AM
P: 123
Hi everyone,

I came across something in my vector calculus textbook (Marsden and Tromba, Edition 5, p. 327) that is confusing me.

"A function f(x,y) is said to be bounded if there is a number M>0 such that -M<=f(x,y)<=M for all (x,y) in the domain of f. A continuous function on a closed rectangle is always bounded, but, for example, f(x,y) = 1/x on (0,1]x[0,1] is continuous but is not bounded, because 1/x becomes arbitrarily large for x near 0. The rectangle (0,1]x[0,1] is not closed, because the endpoint 0 is missing in the first factor."

I do not see how f(x,y) = 1/x could possibly be considered bounded on the closed rectangle [0,1]x[0,1] by the definition given - it still approaches infiniti for x near 0. There is clearly no M for which f(x,y) <= M for all (x,y) in the domain of f, be it on [0,1]x[0,1] or (0,1]x[0,1]. And yet, it is stated, without explanation, that the function is bounded on this rectangle.

I looked at some other sources to try to find a different definition of bounded that might clear up some of the confusion, but to no avail (Wikipedia has an equivalent definition and also says that a continuous function on a closed region is always bound).

Does anyone know what the deal is?

Thanks,

HJ Farnsworth
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Ryker
#2
Feb9-11, 01:22 AM
P: 1,088
f(x,y) = 1/x does not exist on the closed rectangle [0,1]x[0,1], but can only exist on (0,1]x(0,1]. It is namely not defined at 0 (so you cannot just add 0 and close the interval), and, hence, you are right, it is not bounded. Note that even if you did define it piecewise to have a value at 0, then it would not be continuous, since it approaches plus infinity as x goes to zero.
HJ Farnsworth
#3
Feb9-11, 01:45 AM
P: 123
Thank you, I think I understand now.

The definition said every continuous function on a closed rectangle is always bounded. f(x,y)=1/x does not exist at x=0 and cannot be considered continuous at x=0. So as I thought, it is unbounded on the rectangle [0,1]x[0,1].

My mistake was in thinking that the text was implying that f was bounded on the closed rectangle, when in fact it was not, since f is not continuous on that interval and thus need not be bounded by the given definition.

Does that sound correct?

Again, many thanks.

-HJ Farnsworth

Mark44
#4
Feb9-11, 09:52 AM
Mentor
P: 21,307
Simple but frustrating confusion on definition of bounded function

Quote Quote by Ryker View Post
f(x,y) = 1/x does not exist on the closed rectangle [0,1]x[0,1], but can only exist on (0,1]x(0,1]. It is namely not defined at 0 (so you cannot just add 0 and close the interval), and, hence, you are right, it is not bounded. Note that even if you did define it piecewise to have a value at 0, then it would not be continuous, since it approaches plus infinity as x goes to zero.
One slight correction. f is defined on (0, 1] x [0, 1]. Note that 0 is included in the second interval (the domain for y).
Ryker
#5
Feb9-11, 11:51 PM
P: 1,088
Quote Quote by Mark44 View Post
One slight correction. f is defined on (0, 1] x [0, 1]. Note that 0 is included in the second interval (the domain for y).
Sorry, you are correct, of course. I am not familiar with rectangles like this or vector calculus yet, so I just answered what I thought was right for x, but completely forgot about the fact that with y there is no such problem in this specific case


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