Analysis question about continuity and vanishing functions

[diligence]

I can't seem to wrap my head around this concept, I'm hoping you can help me out. Suppose you have a continuous function defined on some compact subset of the plane, say {0 <= x <= 1, 0 <= y <= 1}. I guess the function could be either real or complex valued, but let's just say it's real so we don't have to worry about any funky complex business going on. Also, suppose the function vanishes on the bottom side of the square, {y=0, 0 <= x <= 1}

What is wrong with my following proof that the function must therefore vanish on the entire boundary:

Suppose it does not vanish on {x=0, 0 < y <=1 }. Then there exists an e > 0 such that |f(x,y)| > e for (x,y) in {x=0, 0<y<=1}. Now consider e/2. Continuity implies that as (x,y) approaches the origin along the y axis, there must exist a d>0 such that |(x,y) - (0,0)| < d implies |f(x,y) - f(0,0)| = |f(x,y)| < e/2. But we know that f is bigger than e on this portion of the y-axis, hence this can't be possible. Therefore, f is either not continuous or f also vanishes on the y-axis between 0 and 1. We can then use the same logic to say f vanishes on the entire boundary of the square.

This doesn't feel correct at all, but I can't figure out what's wrong with my proof. Is it because I assumed since f does not vanish that it must be bigger than some e > 0 ?

 

[JG89]

The proposition you're trying to prove is false. But you still made some logical mistakes in your proof:

If you want to show by contradiction that f must vanish entirely on $D = \{(x,y): x = 0, y \in [0,1] \}$ then what you suppose is that there exist a point $(x_0, y_0) \in D$ and an $\epsilon > 0$ such that $|f(x_0, y_0)| > \epsilon$. What you assumed was that $f(x,y) > \epsilon$ for every (x,y) in D.

 

[diligence]

Ah yes that makes sense, thanks. Apparently i need to be more careful.

There's still something I'm a bit unclear about though. If one says that a function is never zero, does that imply there exists an ε > 0 such that |f| > ε everywhere?

 

[HallsofIvy]

Only if it is defined on a compact (in Rⁿ, closed and bounded) set. If not, then f(x)= 1/x is continuous and never 0 on (0, 1) but there is no $\epsilon> 0$ such that $f(x)> \epsilon$ for all x in (0, 1).

 

[blue_raver22]

There are a few issues with your proof. Firstly, you are assuming that the function is bigger than some e > 0 on the y-axis, but this may not necessarily be true. The function could have a value of 0 on the y-axis between 0 and 1, but still be non-zero on the rest of the boundary.

Secondly, your use of the continuity argument is incorrect. While it is true that continuity implies that as (x,y) approaches the origin along the y-axis, there must exist a d>0 such that |(x,y) - (0,0)| < d implies |f(x,y) - f(0,0)| = |f(x,y)| < e/2, this does not necessarily mean that the function must be smaller than e/2 at all points on the y-axis. It only means that there exists a d>0 for a given e/2, but the function could still have values larger than e/2 at other points on the y-axis.

In addition, your argument also does not take into account that the function could have different behavior on different parts of the boundary. Just because it vanishes on the bottom side of the square, does not necessarily mean it must also vanish on the other sides.

To prove that the function vanishes on the entire boundary, you would need to show that it vanishes on all four sides of the square individually, rather than assuming it must vanish on the entire boundary based on one side.

In summary, your proof is not valid because it makes assumptions about the behavior of the function that may not necessarily be true, and it does not consider the possibility of different behavior on different parts of the boundary.