Analysis - Show Linear functions are uniformly Continuous

dkotschessaa
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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. dx(f(P),(Q))<δ→ dy

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.
 
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dkotschessaa said:

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. dx(f(P),(Q))<δ→ dy

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 &gt; 0 there exists a \delta &gt; 0 such that for all x \in U and all y \in U, if |x - y| &lt; \delta then |f(x) - f(y)| &lt; \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)|.
 
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