Help with continuous functions in metric spaces

In summary: Otherwise, we just need to show that if v is any vector in R such that inf||v||=1|f(v)|then v is not in R.Since f is continuous on 0, there is a k such that for each u in v, |f(u)| ≤ k*||u||. So by the inequality you provided, |f(v)| ≤ k*||v||.
  • #1
h20o85
2
0
hi guys,
I have a question I would like assistance with:

let (v,||.||) be a norm space over ℝ, and let f:v→ℝ be a linear functional.
if f is continuous on 0 (by the metric induced by the norm), prove that there is k>0 such that for each u in v, |f(u)| ≤ k*||u||.

thanks :)
 
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  • #2
I am a little confused here, do you consider ℝ also as a normed space, or is f an element of the dual V*? If f is an element of the dual, what notion of continuity do you have?

If f is a linear operator between normed spaces, then continuity at a point implies global continuity (by translation; given T(x), we can get T(y)=T(y-x+x) , and linearity does the rest), and, for linear operators on finite-dimensional space, continuity implies boundedness; express any vector v in terms of a finite basis, and then use inequalities to find a bound for ||T(v)||. Is that the question?
 
  • #3
Thank you for replying (I am new to this forum and I just realized that I am not supposed to ask homework-style questions here...)

R is also considered as a normed space, but v is not necessarily a finite dimensional space, therefore I can't use boundedness...
 
  • #4
h20o85 said:
R is also considered as a normed space, but v is not necessarily a finite dimensional space, therefore I can't use boundedness...

Continuous operators are bounded at the origin even for the infinite dimensional case, but that's essentially what you're proving here so if it hasn't been brought up in class I would be hesitant to use it.

Suppose that the claim is not true. Then for each integer n, there is some un such that |f(u)| > n||u||. Prove f is not continuous at the origin
 
  • #5
h20o85 said:
hi guys,
I have a question I would like assistance with:

let (v,||.||) be a norm space over ℝ, and let f:v→ℝ be a linear functional.
if f is continuous on 0 (by the metric induced by the norm), prove that there is k>0 such that for each u in v, |f(u)| ≤ k*||u||.

thanks :)


We have some δ st for all p with ||p||<δ we have |f(p)-f(0)| = |f(p)| < ε for any ε. Then for any unit vector v in our space

|f(v)| = 1/δ |f(δ v )| < ε||v||/δ =ε/δ since ||δ v || = δ. But the norm of the opearator is

inf||v||=1|f(v)|

so we are done if you just pick for ε any number n .
 
Last edited:

1. What is a continuous function in a metric space?

A continuous function in a metric space is a function that preserves the continuity of a metric. This means that small changes in the input of the function will result in small changes in the output of the function. In simpler terms, a continuous function in a metric space is a function where nearby points in the input space will result in nearby points in the output space.

2. How is continuity defined in a metric space?

In a metric space, continuity is defined as follows: for any point in the input space, there exists a neighborhood around that point such that all points within that neighborhood will be mapped to a neighborhood around the corresponding output point. In other words, the function does not have any sudden jumps or breaks in its mapping from the input space to the output space.

3. What is the importance of continuous functions in metric spaces?

Continuous functions in metric spaces are important because they provide a way to measure the smoothness and connectedness of a function. They also allow us to make predictions and approximations about the behavior of a function in different parts of the metric space. Continuous functions are also the foundation for many mathematical concepts and theories in analysis and topology.

4. How can we determine if a function is continuous in a metric space?

To determine if a function is continuous in a metric space, we can use the epsilon-delta definition of continuity. This definition states that for any epsilon greater than zero, there exists a delta greater than zero such that the distance between the input points is less than delta will result in a distance between the corresponding output points less than epsilon. If this condition is met, then the function is continuous in the metric space.

5. Can a function be continuous in one metric space but not in another?

Yes, a function can be continuous in one metric space but not in another. This is because different metric spaces can have different topologies and structures, which can affect the continuity of a function. For example, a function may be continuous in a discrete metric space, but not in a Euclidean metric space. It is important to consider the specific metric space when determining the continuity of a function.

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