7.4. Local Aspects

As with continuity differentiability at a proves to be local property. It is e.g. compatible with restricting functions to a subset of their domain:

Proposition: Let

a \in A \subset B

be an accumulation point of A (and thus of B) and

f : B \to ℝ

. If f is differentiable at a so is

f | A

with the same deviation:

(f | A)^{'} (a) = f^{'} (a)

[7.4.1]

The reverse is not true in this case as well. Due to an example in 7.3 the absolute value function $| X |$ is not differentiable (at 0) with a differentiable restriction $| X | | ℝ^{\geq 0} = X | ℝ^{\geq 0}$ .

Proof: The assertion is immediate from [6.9.1] when we consider that the difference quotient function of $f | A$ is a restriction of that of f:

m_{f | A, a} = m_{f, a} | A \ {a}

So we have for instance for any subset A of $ℝ$ with an accumulation point $a \in A$ :

(c | A)^{'} (a) = 0

and

(X | A)^{'} (a) = 1

Two functions that coincide locally have the same differentiability behaviour, a feature that also reflects the local character.

Proposition: Let

a \in A \cap B

be an accumulation point of

A \cap B

. If

f : A \to ℝ

and

g : B \to ℝ

coincide locally

at a we have:

f is differentiable at a

\Leftrightarrow

g is differentiable at a.

[7.4.2]

Differentiability assumed we have in addition:

f^{'} (a) = g^{'} (a)

Proof:
As f and g coincide locally at a the same applies to the difference quotient functions $m_{f, a}$ and $m_{g, a}$ . Due to [6.9.2] $m_{f, a}$ is continuously extendable at a iff this is true for $m_{g, a}$ . In that case both difference quotient functions have the same limit.

Consider:

As f and $f | A_{a, ε}$ always coincide locally at a [7.4.2] allows to reduce tests for differentiability to relative neighbourhoods:

f is differentiable at a $\Leftrightarrow f | A_{a, ε}$ is differentiable at a for one $ε > 0$

The derivative numbers, if they exist, are identical: $f^{'} (a) = (f | A_{a, ε})^{'} (a)$ .

Functions defined by sections could be treated niftily using [7.4.2]. We take the absolute value function to demonstrate this.

Example: The absolute value function is differentiable at each $a \neq 0$ . The deviation numbers calculate to

${| X |}^{'} (a) = \frac{| a |}{a} = {\begin{array}{r} 1, if a > 0 \\ - 1, if a < 0 \end{array}$
[7.4.3]

Proof:
For $a > 0$ $| X |$ locally coincides with X, thus yielding the derivative number 1 in this case.
For $a < 0$ $| X |$ locally coincides with −X, thus yielding the derivative number −1 this time.

According to the above mentioned example in 7.3 the absolute value function is not differentiable at 0. [7.4.3] thus proves that it fails to be differentiable at only one point. The next example introduces a function being the exact opposite, i.e. a function which is solely differentiable at a single point. Both examples show that the differential quality of a function at one point has no impact on that quality with the neighbouring points.

Example: The function $f ≔ X^{2} \cdot χ_{ℚ}$ , which means

f (x) = {\begin{array}{l} x^{2}, if x \in ℚ \\ 0, if x \notin ℚ \end{array}

[7.4.4]

is differentiable at 0 but nowhere alse.

Proof:

For each $x \neq 0$ we may estimate as follows

$0 \leq | \frac{f (x) - f (0)}{x - 0} - 0 | = | \frac{f (x)}{x} | \leq | x |$

so that due to [6.9.10]/[6.9.11] f proves to be differentiable at 0 with $f^{'} (0) = 0$ .
If f would be differentiable at some point $a \neq 0$ it would also be continuous at that point (see [7.5.2]) and so would be the function $\frac{f}{X}$ in contrast to [6.2.15].

Another local aspect is the problem to concatenate two functions (cf [6.8.7]), in this case to concatenate them differentiably. The following proposition shows that this is merely the question whether two functions coincide in their values and their derivative numbers at a.

Proposition: Let $f : A \to ℝ$ and $g : B \to ℝ$ be any two functions such that $f | A \cap B = g | A \cap B$ and let $a \in A \cap B$ be an accumulation point of A and of B as well. Then the following equivalence holds:

	$f \cup g$ is differentiable at a
$\Leftrightarrow$	f and g are both differentiable at a and $f^{'} (a) = g^{'} (a)$

[7.4.5]

Proof: Considering the identities $f = f \cup g | A$ and $g = f \cup g | B$ the direction " $\Rightarrow$ " is immediate consequence of [7.4.1].

" $\Leftarrow$ ": Due to the premise we have for all $x \in A \cap B, x \neq a$

\frac{f (x) - f (a)}{x - a} = \frac{g (x) - g (a)}{x - a}

so that $m_{f, a} | A \cap B = m_{g, a} | A \cap B$ . Now, as $\lim_{x \to a} m_{f, a} (x) = f^{'} (a) = g^{'} (a) = \lim_{x \to a} m_{g, a} (x)$ , [6.8.9] guarantees that the function

m_{f \cup g, a} = m_{f, a} \cup m_{g, a}

is continuously extentable at a. But that means: $f \cup g$ is differentiable at a with $f^{'} (a) = g^{'} (a)$ as derivative number.

7.3. 7.5.