The Unique Limit of a Complex Function

[Calu]

Homework Statement

I'm struggling with the proof that the limit of a complex function is unique. I'm struggling to see how |L-f(z*)| + |f(z*) - l'| < ε + ε is obtained.

Homework Equations

0 < |z-z₀| < δ implies |f(z) - L| < ε, where L is the limit of f(z) as z→z₀ .

The Attempt at a Solution

Let S ⊆ ℂ.

We consider some L' ≠ L, a limit of the function f(z) as z→z₀. Choose ε = 1/2|L-L'| to find δ₁ > 0, δ₂ > 0 such that

z ∈ S, 0 < |z-z₀| < δ₁ implies |f(z) - L| < ε
z ∈ S, 0 < |z-z₀| < δ₂ implies |f(z) - L'| < ε

Note that z₀ is a limit point, so there exists some z* ∈ S such that 0 < |z₀ - z*| < min {δ₁, δ₂}. Then

|L-L'| = |L-f(z*) + f(z*) - L'| ≤ |L - f(z*)| + |f(z*) - L'| < ε + ε.

This is the part where I get confused. I don't see how we can say that |L - f(z*)| + |f(z*)- L'| < ε + ε.

 

[Mark44]

Homework Statement

I'm struggling with the proof that the limit of a complex function is unique. I'm struggling to see how |L-f(z*)| + |f(z*) - l'| < ε + ε is obtained.

Homework Equations

0 < |z-z₀| < δ implies |f(z) - L| < ε, where L is the limit of f(z) as z→z₀ .

The Attempt at a Solution

Let S ⊆ ℂ.

We consider some L' ≠ L, a limit of the function f(z) as z→z₀. Choose ε = 1/2|L-L'| to find δ₁ > 0, δ₂ > 0 such that

z ∈ S, 0 < |z-z₀| < δ₁ implies |f(z) - L| < ε
z ∈ S, 0 < |z-z₀| < δ₂ implies |f(z) - L'| < ε

Note that z₀ is a limit point, so there exists some z* ∈ S such that 0 < |z₀ - z*| < min {δ₁, δ₂}. Then

|L-L'| = |L-f(z*) + f(z*) - L'| ≤ |L - f(z*)| + |f(z*) - L'| < ε + ε.

This is the part where I get confused. I don't see how we can say that |L - f(z*)| + |f(z*)- L'| < ε + ε.

Both |L - f(z*)| and |f(z*) - L'| are smaller than ε, so their sum would be smaller than 2ε. Is that the part you're confused on?
Or are you confused about how they went from |L-f(z*) + f(z*) - L'| to |L - f(z*)| + |f(z*) - L'|?

 

[Calu]

Both |L - f(z*)| and |f(z*) - L'| are smaller than ε, so their sum would be smaller than 2ε. Is that the part you're confused on?
Or are you confused about how they went from |L-f(z*) + f(z*) - L'| to |L - f(z*)| + |f(z*) - L'|?

That is the part I'm confused on. I realize that they must be smaller than ε, I just don't see how they are.

 

[Mark44]

That is the part I'm confused on. I realize that they must be smaller than ε, I just don't see how they are.

Since both L and L' are assumed to be the limits, it has to be the case that |f(z*) - L| < ε and that |f(z*) - L'| < ε. Most of the limit proofs work backwards from this conclusion to determine what the δ needs to be.

 

[Calu]

Which one? I listed two of the points where I though you might be having difficulties.

Sorry, I thought I'd deleted part of the quote. This is the part I'm confused about:

Both |L - f(z*)| and |f(z*) - L'| are smaller than ε, so their sum would be smaller than 2ε.

I realize that they must be smaller than ε, I just don't see how they are.

 

[Mark44]

See my edited post. I changed it after I wrote it.

 

[Calu]

I think this is where I'm confused: In the first part there is |L - f(z*)| < ε, whereas you've written that is has to be the case that |f(z*) - L| < ε. Are the two equivalent? I see now how |f(z*) - l'| < ε and |f(z*) - l'| < ε must be true as they come from the definition of the limit. However, we have |L - f(z*)|.

 

[Mark44]

I think this is where I'm confused: In the first part there is |L - f(z*)| < ε, whereas you've written that is has to be the case that |f(z*) - L| < ε. Are the two equivalent?

The two are equal.
It's always the case that |a - b| = |b - a|.

I see now how |f(z*) - l'| < ε and |f(z*) - l'| < ε must be true as they come from the definition of the limit. However, we have |L - f(z*)|.

 

[Calu]

The two are equal.
It's always the case that |a - b| = |b - a|.

Thanks very much for your help.