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 [itex]P[/itex], [itex]Q[/itex], [itex]d_x[/itex] and [itex]d_y[/itex]?

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

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

Now calculate [itex]|f(x) - f(y)|[/itex].