Linear function F continuous somewhere, to prove continuous everywhere

[SrEstroncio]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

I tried using several forms of the triangle inequality to prove that \|f(\vec{x})\|<\epsilon implies that \|f(\vec{x})-f(\vec{x_0})\|<\epsilon by means of adding a zero f(x)=f(x+0)=f(x+x_0-x_0)=f(x)-f(x_0)+f(x_0) but I haven't been able to conclude anything special.

Thanks in advance for all your help.

 

[hunt_mat]

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...

 

[Office_Shredder]

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)

 

[SrEstroncio]

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.

 

[Office_Shredder]

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

 

[SrEstroncio]

Got it. Thanks