# Analysis - Show Linear functions are uniformly Continuous

1. Jan 12, 2014

### dkotschessaa

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

Suppose f:R->R is a linear function. Prove from the definition that f is uniformly continuous on R.

2. Relevant equations

Epsilon delta definition of uniform continuity: A function f:X->Y is called uniformly continuous if $\forall\epsilon$>0 ∃x st. dx(f(P),(Q))<δ→ dy

3. The attempt at a solution

I found this easier than I expected, so of course that makes me think I'm wrong. Also I'm not sure about the placement of the absolute value signs near the end.

let σ=f-1(ε)

|p-q|<σ→|p-q|<f-1ε → f(|p-q|)≤|f(p)-f(q)| < ε

Where f(|p-q|)≤|f(p)-f(q)| is due to the linearity of f.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 12, 2014

### pasmith

What are $P$, $Q$, $d_x$ and $d_y$?

The definition of uniform continuity for real functions is that $f$ is uniformly continuous on $U \subset \mathbb{R}$ if and only if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x \in U$ and all $y \in U$, if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$.

This step is not justified, since constant functions are linear but not invertible.

If $f: \mathbb{R} \to \mathbb{R}$ is linear, then you must have $f: x \mapsto ax + b$ for some $a \in \mathbb{R}$ and $b \in \mathbb{R}$.

Now calculate $|f(x) - f(y)|$.