Analysis - Show Linear functions are uniformly Continuous

[dkotschessaa]

Homework Statement

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

Homework Equations

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

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.

 

[pasmith]

Homework Statement

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

Homework Equations

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

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 $|f(x) - f(y)| < \epsilon$.

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)

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