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8. DIFFERENTIATION OF REAL VALUED FUNCTIONS

Much of the elementary theory of differential calculus rests on a few simple properties of the families $\mathfrak{O}$ and $\mathrm{0}$. These are given in propositions

8.1.6. Definition. A function $\mathit{L}\mathrm{:}\mathbb{R}\mathrm{\to }\mathbb{R}$ is linear if

$\mathit{L}\mathrm{\left(}\mathit{x}\mathrm{+}\mathit{y}\mathrm{\right)}\mathrm{=}\mathit{L}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{+}\mathit{L}\mathrm{\left(}\mathit{y}\mathrm{\right)}$

and

$\mathit{L}\mathrm{\left(}\mathit{c}\mathit{x}\mathrm{\right)}\mathrm{=}\mathit{c}\mathit{L}\mathrm{\left(}\mathit{x}\mathrm{\right)}$

for all $\mathit{x}\mathrm{,}$ $\mathit{y}\mathrm{,}$ $\mathit{c}\mathrm{\in }\mathbb{R}$. The family of all linear functions from $\mathbb{R}$ into $\mathbb{R}$ will be denoted by $\mathit{L}\mathrm{.}$

The collection of linear functions from $\mathbb{R}$ into $\mathbb{R}$ is not very impressive, as the next problem shows. When we get to spaces of higher dimension the situation will become more interesting.

8.1.7. Example. A function $\mathit{f}\mathrm{:}\mathbb{R}\mathrm{\to }\mathbb{R}$ is linear if and only if its graph is $\mathrm{a}$ (nonvertical) line through the origin.

Proof. Problem.

CAUTION. Since linear functions must pass through the origin, straight lines are not in general graphs of linear functions.

8.1.8. Proposition. Every member of $\mathrm{0}$ belongs to $\mathfrak{O}$; so does every member of L. Every member of $\mathfrak{O}$ is continuous at $\mathit{0}\mathrm{.}$

Proof. Obvious from the definitions. $\mathrm{\square }$

8.1.9. Proposition. Other than the constant function zero, no linear function belongs to $\mathrm{0}\mathrm{.}$

Proof. Exercise. (Solution .)

8.1.10. Proposition. The family $\mathfrak{O}$ is closed under addition and multiplication by constants.

Proof. Exercise. (Solution .)

8.1.11. Proposition. The fa mily $\mathrm{0}$ is closed under addition and multiplication by constants.

Proof. Problem.

The next two propositions say that the composite of a function in $\mathfrak{O}$ with one in $\mathrm{0}$ (in either order) ends up in $\mathrm{0}\mathrm{.}$

8.1.12. Proposition. If $\mathit{g}\mathrm{\in }\mathfrak{O}$ and $\mathit{f}\mathrm{\in }\mathrm{0}$, then $\mathit{f}\mathrm{o}\mathit{g}\mathrm{\in }\mathrm{0}\mathrm{.}$

Proof. Problem.

8.1.13. Proposition. If $\mathit{g}\mathrm{\in }\mathrm{0}$ and $\mathit{f}\mathrm{\in }\mathfrak{O}$, then $\mathit{f}\mathrm{\circ }\mathit{g}\mathrm{\in }\mathrm{0}\mathrm{.}$

Proof. Exercise. (Solution .)

8.1.14. Proposition. If $\mathit{\phi }\mathrm{,}$ $\mathit{f}\mathrm{\in }\mathfrak{O}$, then $\mathit{\phi }\mathit{f}\mathrm{\in }\mathrm{0}\mathrm{.}$

Proof. Exercise. (Solution .)

Remark. The preceding facts can be summarized rather concisely. (Notation: ${\mathit{C}}_{\mathrm{0}}$ is the set of all functions in ${{F}}_{\mathrm{0}}$ which are continuous at $\mathrm{0}\mathrm{.}$)

(1) $\mathit{L}\mathrm{\cup }\mathrm{0}\mathrm{\subseteq }\mathfrak{O}\mathrm{\subseteq }{\mathit{C}}_{\mathrm{0}}\mathrm{.}$

(2) $\mathit{L}\mathrm{\cap }\mathrm{0}\mathrm{=}\mathrm{0}\mathrm{.}$

(3) $\mathfrak{O}\mathrm{+}\mathfrak{O}\mathrm{\subseteq }\mathfrak{O}$; $\mathit{\alpha }\mathfrak{O}\mathrm{\subseteq }\mathfrak{O}\mathrm{.}$

(4) $\mathrm{0}\mathrm{+}\mathrm{0}\mathrm{\subseteq }\mathrm{0}$; $\mathit{\alpha }\mathrm{0}\mathrm{\subseteq }\mathrm{0}\mathrm{.}$

(5) $\mathrm{0}\mathrm{\circ }\mathfrak{O}\mathrm{\subseteq }\mathrm{0}\mathrm{.}$

(6) $\mathfrak{O}\mathrm{\circ }\mathrm{0}\mathrm{\subseteq }\mathrm{0}\mathrm{.}$

(7) $\mathfrak{O}\mathrm{\cdot }\mathfrak{O}\mathrm{\subseteq }\mathrm{0}\mathrm{.}$