What Conditions Ensure Convergence of Sequences?

[DespotDespond]

Hi,

I have a basic question about convergence.

I have two sequences, x₁, x₂, ... and y₁, y₂, ..., where y_n = f(x_n) for some function f : ℝ^N → ℝ.

I have shown that the sequence, y₁, y₂, ... converges. What conditions do I need on the function, f, to ensure that the sequence x₁, x₂, ... also converges?

Thanks in advance

 

[lurflurf]

You need the function f to be continuous.

 

[Office_Shredder]

In fact this is an equivalent definition of being continuous: f is continuous if and only if whenever $x_n \to x$, $f(x_n) \to f(x)$.

 

[R136a1]

But the poster is asking the converse. He's asking if $f(x_n)\rightarrow x$ implies $x_n\rightarrow x$. So continuity is irrelevant here.

 

[DespotDespond]

Indeed, R136a1 is correct. I want the convergence of the x_n → x given the convergence of f(x_n) → f(x).

In this case, continuity of f is not enough. For example, f(x) = x² is continuous, but if we define the two series as follows

x₁ = √2, x₂ = -√2, x₃ = √2, x₄ = -√2, ...

and

y₁ = 2, y₂ = 2, y₃ = 2, y₄ = 2, ...

then y_n is convergent and y_n = f(x_n), but x_n is not convergent.

If the function, f, were invertible and the inverse was continuous, then that would be enough. Right?

The problem is that in my case I don't think that f is invertible.

I don't necessarily need the result to hold globally, i.e. for all x. A local result might suffice.

Any syggestions would be appreciated.

 

[R136a1]

Indeed, R136a1 is correct. I want the convergence of the x_n → x given the convergence of f(x_n) → f(x).

In this case, continuity of f is not enough. For example, f(x) = x² is continuous, but if we define the two series as follows

x₁ = √2, x₂ = -√2, x₃ = √2, x₄ = -√2, ...

and

y₁ = 2, y₂ = 2, y₃ = 2, y₄ = 2, ...

then y_n is convergent and y_n = f(x_n), but x_n is not convergent.

If the function, f, were invertible and the inverse was continuous, then that would be enough. Right?

The problem is that in my case I don't think that f is invertible.

I don't necessarily need the result to hold globally, i.e. for all x. A local result might suffice.

Any syggestions would be appreciated.

You don't even need fully invertible. Just a continuous left-inverse suffices.

I kind of doubt there is a simple condition. Could you perhaps give some more information about your specific situation?

 

[DespotDespond]

Hi,

Thanks for the response.

The problem is related to an algorithm that I've constructed for optimising Markov decision processes. I'm trying to prove the global convergence of the algorithm. I don't want to go into any unnecessary detail, so I will try to explain as briefly as I can.

I have an objective function, f, of which the x's are the parameters I wish to optimise.

My algorithm generates a series of parameter vectors, x₁, x₂, ...

I've managed to show that the objective is strictly monotonically increasing with respect to this series, i.e. that

f(x₁) < f(x₂) < ...

I know that f is bounded from above, so I used the monotone convergence theorem to conclude that the series

f(x₁), f(x₂), ...

converges.

Generally, f is not injective and so doesn't have an inverse.

 

[R136a1]

Thanks for your response. But what I really wanted to know is whether $f$ has some special properties that we could use. We know it's not injective, but is there perhaps something else? Obviously, if $f$ can be any continuous function, then what you're trying to do is false. So there must be something special going on.