Linear Function is Continuous

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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

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