# Help with continuous functions in metric spaces

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 :)

Bacle2
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?

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...

Office_Shredder
Staff Emeritus
Gold Member
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

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: