Continuity of $f(x)$ at 0 Using $g_r(x)$

[vintwc]

Homework Statement

If $f:\mathbb{R} \to \mathbb{R}$ is such that for all $r>0$ there exists a continuous function $g_r \mathbb{R} \to \mathbb{R}$ such that $|g_r (x) - f(x)| < r$ for $|x| < 1$ then f is continuous at 0.

Homework Equations

The Attempt at a Solution

When $|x| < \delta _g$, $|g_r (x) - g_r (0)| < \epsilon '$ ...(1)
When $|x| < 1$, $|g_r (x) - f(x)| < r$ , i.e. $|f(x) - g_r (x)| < r$ ...(2)
$|g_r(0) - f(0)| < r$ ...(3)

Adding (1) and (2) gives $|f(x) - g_r (0)| < \epsilon ' + r$ ...(4)
Adding (3) and (4) gives $|f(x) - f(0)| < \epsilon ' + 2r$
So when |x| is the $\min\{1, \delta_g\}$ , $|f(x) - f(0)| < \epsilon$
And by the previous result, we can say that f is continuous at 0.

Not sure whether its right for me to take the $\delta = \min\{1, \delta_g\}$

 

[lanedance]

I think it is ok, though it doesn't make much sense to have delta greater than 1 anyway...