By definition, a affine function f:Rn→Rmf : R^n\rightarrow R^m is affine if

f(tx+(1âˆ′t)y)=tf(x)+(1âˆ′t)f(y)f(tx + (1 âˆ’ t)y) = t f(x) + (1 âˆ’ t) f(y) for all x, y \in R^nx, y \in R^n and t \in [0, 1]t \in [0, 1]. However, we can also write ff in the form f(x)=Ax+bf(x)=Ax+b where AA is a constant m أ— nm أ— n matrix and bb is a constant mm-vector.

Recall also that a function f : A â†’ Rf : A â†’ R is uniformly continuous on AA if for

every \epsilon > 0\epsilon > 0 there exists a \delta > 0\delta > 0 such that |x âˆ’ y| < خ´|x âˆ’ y| < خ´ implies |f(x) âˆ’ f(y)| < \epsilon|f(x) âˆ’ f(y)| < \epsilon.
\mathbf Claim\mathbf Claim: An affine function is uniformly continuous.
Proof:
Using the definition of an affine function in the form f(x)=Ax+bf(x)=Ax+b, we have that
\vert f(x) - f(y) \vert = \vert Ax+b-Ay-b\vert = \vert A(x-y)\vert\le\vert A\vert \vert x-y\vert\vert f(x) - f(y) \vert = \vert Ax+b-Ay-b\vert = \vert A(x-y)\vert\le\vert A\vert \vert x-y\vert
Thus, for any \epsilon>0\epsilon>0, if we take \delta =\epsilon/\vert A \vert\delta =\epsilon/\vert A \vert, this implies that

\vert f(x) – f(y) \vert \le \vert A\vert \vert x-y\vert< \epsilon \vert f(x) - f(y) \vert \le \vert A\vert \vert x-y\vert< \epsilon
Thus, ff is uniformly continuous.
Comments:
There are several issues that I am not sure with this proof.
I am assuming f(x)f(x) and f(y)f(y) have the same matrix AA and vector bb, but this was the only way I could figure out in order to get to the expression \vert A\vert \vert x-y\vert\vert A\vert \vert x-y\vert and then use the |x âˆ’ y| < خ´|x âˆ’ y| < خ´ argument.
Also, I am taking the expression \delta =\epsilon/\vert A \vert\delta =\epsilon/\vert A \vert, which is depending on the matrix AA itself and as far as I know, when we talk about uniform converge, our choice of \delta\delta must not be depending on something else.
What other problems does the proof have? How could I form a better proof?
=================
There is a typo in the first line of equations after proof: it should have been \lvert A(x-y)\rvert \le \lvert A\rvert \lvert x-y\rvert\lvert A(x-y)\rvert \le \lvert A\rvert \lvert x-y\rvert As for your question: of course you take the same ff and only one AA depending on ff, because you are trying to prove that the single function f(x):=Ax+bf(x):=Ax+b is uniformly continuous, not that the function g:\mathcal M_{m\times n}(\Bbb R)\times \Bbb R^m\times \Bbb R^n\to \Bbb R^n \\ g(A,b,x)=Ax+bg:\mathcal M_{m\times n}(\Bbb R)\times \Bbb R^m\times \Bbb R^n\to \Bbb R^n \\ g(A,b,x)=Ax+b is uniformly continuous (and, in fact, it is not, because it has a bilinear part).
– G. Sassatelli
Oct 20 at 20:17
=================
1 Answer
1
=================
1.) Of course, it is the same function ff and AA and bb are constants of ff.
2.) AA is, again, a constant of ff, the variable is xx, and only that variable vector is relevant in the uniform continuity definition.