Linear function F continuous somewhere, to prove continuous everywhere

  • #1

Homework Statement


Let [tex] f:A\subset{\mathbb{R}}^{n}\mapsto \mathbb{R} [/tex] be a linear function continuous a [tex] \vec{0} [/tex]. To prove that [tex]f[/tex] is continuous everywhere.


Homework Equations


If [tex] f[/tex] is continuous at zero, then [tex]\forall \epsilon>0 \exists\delta>0 [/tex] such that if [tex] \|\vec{x}\|<\delta[/tex] then [tex] \|f(\vec{x}) \|< \epsilon [/tex].
[tex] f[/tex] also satisfies [tex] f(\vec{a}+\alpha\vec{b})=f(\vec{a})+\alpha f(\vec{b}) [/tex].

The Attempt at a Solution


I tried using several forms of the triangle inequality to prove that [tex] \|f(\vec{x})\|<\epsilon [/tex] implies that [tex] \|f(\vec{x})-f(\vec{x_0})\|<\epsilon [/tex] by means of adding a zero [tex] f(x)=f(x+0)=f(x+x_0-x_0)=f(x)-f(x_0)+f(x_0) [/tex] but I havent been able to conclude anything special.

Thanks in advance for all your help.
 

Answers and Replies

  • #2
hunt_mat
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If you examine the value of the functional at the point ax then you will see that as f is continuous at 0, you can still make f(ax) as small as you like in the norm. Choose a point x_{0} and examine f(x-x_{0}), you know that ||f(x)||<epsilon for all values of x, so...
 
  • #3
Office_Shredder
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You don't know that f(x) is going to be small (in general it won't be).

Try just using linearity: f(x)-f(x0)=f(x-x0)
 
  • #4
I don't seem to be catching the drift. I can't figure out how to use the linearity property in order to get to where I want to be.
 
  • #5
Office_Shredder
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Here's the general idea and you can fill in the details:

We know that if y is small, f(y) is small by continuity at 0. You want to show f(x)-f(x0) is small. But we know that x-x0 is small which means f(x-x0) is small
 
  • #6
Got it. Thanks
 

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