# Linear function F continuous somewhere, to prove continuous everywhere

1. Sep 12, 2010

### SrEstroncio

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 $$f(\vec{a}+\alpha\vec{b})=f(\vec{a})+\alpha f(\vec{b})$$.

3. 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 havent been able to conclude anything special.

2. Sep 12, 2010

### 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...

3. Sep 12, 2010

### Office_Shredder

Staff Emeritus
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. Sep 12, 2010

### 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.

5. Sep 12, 2010

### Office_Shredder

Staff Emeritus
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. Sep 12, 2010

### SrEstroncio

Got it. Thanks