# 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:

 $\left(f|A{\right)}^{\prime }\left(a\right)={f}^{\prime }\left(a\right)$ [7.4.1]

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

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:

${m}_{f|A,a}={m}_{f,a}|A\\left\{a\right\}$

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

$\left(c|A{\right)}^{\prime }\left(a\right)=0$  and   $\left(X|A{\right)}^{\prime }\left(a\right)=1$

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 i i.e. there are two relative ε-neighbourhoods ${A}_{a,\epsilon }=A\cap \right]a-\epsilon ,a+\epsilon \left[$${B}_{a,\epsilon }=B\cap \right]a-\epsilon ,a+\epsilon \left[$ such that ${A}_{a,\epsilon }={B}_{a,\epsilon }$ and $f\left(x\right)=g\left(x\right)$ for all $x\in {A}_{a,\epsilon }\phantom{\rule{1pt}{0ex}}$.
at a we have:

 f is differentiable at a $⇔$ g is differentiable at a. [7.4.2]

Differentiability assumed we have in addition:  ${f\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)={g\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)$.

Proof:
As  f and g coincide locally at a the same applies to the difference quotient functions ${m}_{\phantom{\rule{0.0em}{0ex}}f,a}$ and ${m}_{\phantom{\rule{0.0em}{0ex}}g,a}$. Due to [6.9.2]  ${m}_{\phantom{\rule{0.0em}{0ex}}f,a}$ is continuously extendable at a iff this is true for ${m}_{\phantom{\rule{0.0em}{0ex}}g,a}$. In that case both difference quotient functions have the same limit.

Consider:

• As  f and  $f|{A}_{a,\epsilon }$ always coincide locally at a [7.4.2] allows to reduce tests for differentiability to relative neighbourhoods:

f  is differentiable at a $\text{ }⇔\text{ }f|{A}_{a,\epsilon }$  is differentiable at a for one $\epsilon >0$

The derivative numbers, if they exist, are identical:  ${f}^{\prime }\left(a\right)=\left(f|{A}_{a,\epsilon }{\right)}^{\prime }\left(a\right)$.

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 $a\ne 0$ . The deviation numbers calculate to

 [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  $f≔{\mathrm{X}}^{2}\cdot {\mathrm{\chi }}_{ℚ}$ , which means

 [7.4.4]

is differentiable at 0 but nowhere alse.

Proof:

• For each $x\ne 0$ we may estimate as follows

$0\le |\frac{f\left(x\right)-f\left(0\right)}{x-0}-0|=|\frac{f\left(x\right)}{x}|\le |x|$

so that due to [6.9.10]/[6.9.11]  f proves to be differentiable at 0 with  ${f}^{\prime }\left(0\right)=0$.

• If  f would be differentiable at some point $a\ne 0$ it would also be continuous at that point (see [7.5.2]) and so would be the function $\frac{f}{\mathrm{X}}$ 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 $⇔\text{ }$ f and g are both differentiable at a and  ${f}^{\prime }\left(a\right)={g}^{\prime }\left(a\right)$
[7.4.5]

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

"$⇐$":  Due to the premise we have for all $x\in A\cap B,\text{\hspace{0.28em}}x\ne a$

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

so that ${m}_{f,a}|A\cap B={m}_{g,a}|A\cap B$. Now, as $\underset{x\to a}{\mathrm{lim}}{m}_{f,a}\left(x\right)={f}^{\prime }\left(a\right)={g}^{\prime }\left(a\right)=\underset{x\to a}{\mathrm{lim}}{m}_{g,a}\left(x\right)$, [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  ${f}^{\prime }\left(a\right)={g}^{\prime }\left(a\right)$ as derivative number. 