Affine function is uniform continuous By definition, a affine function $f : R^n\rightarrow R^m$ is affine if
$f(tx + (1 − t)y) = t f(x) + (1 − t) f(y)$ for all $x, y \in R^n$ and $t \in [0, 1]$. However, we can also write $f$ in the form $f(x)=Ax+b$ where $A$ is a constant $m × n$ matrix and $b$ is a constant $m$-vector. 
Recall also that a function $f : A → R$ is uniformly continuous on $A$ if for
every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x − y| < δ$ implies $|f(x) − f(y)| < \epsilon$.
$\mathbf Claim$: An affine function is uniformly continuous. 
Proof:
Using the definition of an affine function in the form $f(x)=Ax+b$, we have that
$$\vert f(x) - f(y) \vert = \vert Ax+b-Ay-b\vert = \vert A(x-y)\vert\le\vert A\vert \vert x-y\vert$$
Thus, for any $\epsilon>0$, if we take $\delta =\epsilon/\vert A \vert$, this implies that 
$$\vert f(x) - f(y) \vert \le \vert A\vert \vert x-y\vert< \epsilon $$
Thus, $f$ is uniformly continuous.
Comments:
There are several issues that I am not sure with this proof.


*

*I am assuming $f(x)$ and $f(y)$ have the same matrix $A$ and vector $b$, but this was the only way I could figure out in order to get to the expression  $\vert A\vert \vert x-y\vert$ and then use the $|x − y| < δ$ argument.

*Also, I am taking the expression $\delta =\epsilon/\vert A \vert$, which is depending on the matrix $A$ itself and as far as I know, when we talk about uniform converge, our choice of $\delta$ must not be depending on something else.
What other problems does the proof have? How could I form a better proof?

1.) Of course, it is the same function $f$ and $A$ and $b$ are constants of $f$.
2.) $A$ is, again, a constant of $f$, the variable is $x$, and only that variable vector is relevant in the uniform continuity definition.
