Bounded Functions Homework: Rudin's R^2

[WackStr]

Homework Statement

This is from baby rudin:

If $$E\subset X$$ and if f is a function defined on X, the restriction of f to E is the function g whose domain of definition is E, such that $$g(p)=f(p)$$ for $$p\in E$$. Define f and g on R^2 by: $$f(0,0)=g(0,0)=0$$, $$f(x,y)=\frac{xy^2}{x^2+y^4}, g(x,y)=\frac{xy^2}{x^2+y^6}$$ if $$(x,y)\neq (0,0)$$. Prove that f is bounded on R^2, that g is unbounded in every neighborhood of $$(0,0)$$, and that f is not continuous at (0,0); nevertheless, the restrictions of both f and g to every straight line in R^2 are continuous.

Homework Equations

In the question statement

The Attempt at a Solution

This is after the continuity chapter and before the differentiation chapter, so I am clueless here. I was thinking of bounding the function by another function but didn't get anywhere. Anyone want to help?

Thanks,

 

[mighty2000]

it is important to approach problems with a systematic and logical approach. In this case, we are asked to prove that f is bounded on R^2, g is unbounded in every neighborhood of (0,0), and that f is not continuous at (0,0).

First, let's define what it means for a function to be bounded. A function f is bounded on a set S if there exists a real number M such that |f(x)| ≤ M for all x in S.

Now, let's consider the function f(x,y) = xy^2 / (x^2 + y^4). We can see that whenever x and y are not equal to 0, the function is well-defined and finite. However, when x = 0, the function becomes undefined. This means that f is not defined at (0,0). Therefore, we cannot say anything about its boundedness at (0,0).

Next, let's consider the function g(x,y) = xy^2 / (x^2 + y^6). We can see that for any neighborhood of (0,0), there exists a point (x,y) within that neighborhood where the denominator x^2 + y^6 is very small, approaching 0. This makes the value of g(x,y) very large, approaching infinity. Therefore, g is unbounded in every neighborhood of (0,0).

Lastly, let's consider the continuity of f and g at (0,0). In order for a function to be continuous at a point, it must exist at that point and the limit of the function as it approaches that point must exist and be equal to the function value at that point. However, as we saw earlier, f is not defined at (0,0). This means that it cannot be continuous at (0,0). On the other hand, the restrictions of both f and g to every straight line in R^2 are continuous as the limit of the function as it approaches any point on a straight line will exist and be equal to the function value at that point.

In conclusion, we have proven that f is bounded on R^2, g is unbounded in every neighborhood of (0,0), and that f is not continuous at (0,0). We have also shown that the restrictions of both f and g to every straight line in R^2 are continuous.