Linear Function is Continuous

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\alpha, \beta \in \R$ be real numbers.

Let $f$ be the real function defined as $f \left({x}\right) = \alpha x + \beta$.


Then $f$ is continuous at every real number $c \in \R$.


Proof

First assume $\alpha \ne 0$.

Let $\epsilon > 0$.

Let $\delta = \dfrac {\epsilon}{\left\vert\alpha\right\vert}$.

Then, provided that $\left\vert{x - c}\right\vert < \delta$:

\(\displaystyle \left\vert{f \left({x}\right) - f \left({c}\right)}\right\vert\) \(=\) \(\displaystyle \left\vert{\alpha \left({x - c}\right)}\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{\alpha}\right\vert \cdot \left\vert{x - c}\right\vert\)
\(\displaystyle \) \(<\) \(\displaystyle \left\vert{\alpha}\right\vert \delta\)
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)

So, we have found a $\delta$ for a given $\epsilon$ so as to make $\left\vert{f \left({x}\right) - f \left({c}\right)}\right\vert < \epsilon$ provided $\left\vert{x - c}\right\vert < \delta$.

So $\displaystyle \lim_{x \to c} f \left({x}\right) = f \left({c}\right)$ and so $f$ is continuous at $c$, whatever $c$ happens to be.


Now suppose $\alpha = 0$.

Then $\forall x \in \R: f \left({x}\right) - f \left({c}\right) = 0$.

So whatever $\epsilon > 0$ we care to choose, $\left\vert{f \left({x}\right) - f \left({c}\right)}\right\vert < \epsilon$, and whatever $\delta$ may happen to be is irrelevant.

Continuity follows for all $c \in \R$, as above.

$\blacksquare$


Sources