Linear Function is Continuous
Theorem
Let $\alpha, \beta \in \R$ be real numbers.
Let $f : \R \to \R$ be a linear real function:
- $\map f x = \alpha x + \beta$
for all $x \in \R$.
Then $f$ is continuous at every real number $c \in \R$.
Proof 1
First assume $\alpha \ne 0$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \dfrac \epsilon {\size \alpha}$.
Then, provided that $\size {x - c} < \delta$:
\(\ds \size {\map f x - \map f c}\) | \(=\) | \(\ds \size {\alpha \paren {x - c} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {\alpha} \cdot \size {x - c}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \size {\alpha} \delta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
So, we have found a $\delta$ for a given $\epsilon$ so as to make $\size {\map f x - \map f c} < \epsilon$ provided $\size {x - c} < \delta$.
So $\ds \lim_{x \mathop \to c} \map f x = \map f c$ and so $f$ is continuous at $c$, whatever $c$ happens to be.
Now suppose $\alpha = 0$.
Then $\forall x \in \R: \map f x - \map f c = 0$.
So whatever $\epsilon \in \R_{>0}$ we care to choose, $\size {\map f x - \map f c} < \epsilon$, and whatever $\delta$ may happen to be is irrelevant.
Continuity follows for all $c \in \R$, as above.
$\blacksquare$
Proof 2
Let $c \in \R$.
Let $\sequence {x_n}$ be a real sequence converging to $c$.
Then:
\(\ds \lim_{n \mathop \to \infty} \map f {x_n}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\alpha x_n + \beta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha c + \beta\) | Combined Sum Rule for Real Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f c\) |
We therefore have:
- for all real sequences $\sequence {x_n}$ converging to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
So by Sequential Continuity is Equivalent to Continuity in the Reals $f$ is continuous at $c$.
As $c \in \R$ was arbitrary, $f$ is continuous on $\R$.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 8.7$