A function *f*:(*a*, *b*) → ℝ may, in general, show a quite pathological behavior in the neighborhood of one of its zeros (see, e.g., Examples 2.2.3 and 2.9 below). To exclude such exotic cases but still be sufficiently general to cover most of the relevant cases, we use the following definition.

#### Definition 2.1

A zero *x*_{0} ∈ (*a*, *b*) of a function *f* ∈ *C*^{0}(*a*, *b*) ∩ *C*^{1}((*a*, *b*)∖{*x*_{0}}) will be called *admissible* provided

$$\begin{array}{}{\displaystyle \underset{x\nearrow {x}_{0}}{lim}\frac{{f}^{\prime}(x)}{f(x)}=-\mathrm{\infty}\text{\hspace{0.17em}and\hspace{0.17em}}\underset{x\searrow {x}_{0}}{lim}\frac{{f}^{\prime}(x)}{f(x)}=\mathrm{\infty}.}\end{array}$$(2.1)

If *f* extends continuously to *a* (or *b*) and *f*(*a*) = 0 (or *f*(*b*) = 0), we will say that *f* has an *admissible zero* in *a* (or *b*) if

$$\begin{array}{}{\displaystyle \underset{x\searrow a}{lim}\frac{{f}^{\prime}(x)}{f(x)}=\mathrm{\infty}\text{\hspace{0.17em}}\left(\text{or\hspace{0.17em}}\underset{x\nearrow b}{lim}\frac{{f}^{\prime}(x)}{f(x)}=-\mathrm{\infty}\right).}\end{array}$$

#### Example 2.2

The function *f*_{1} ∈ *C*^{0}(ℝ) ∩ *C*^{∞}(ℝ∖{0}), *x* ↦
$\begin{array}{}\sqrt{|x|}\end{array}$
has an admissible zero in *x* = 0 (see Remark 6 above).

The *C*^{∞}-function

$$\begin{array}{}{\displaystyle {f}_{2}(x):=\left\{\begin{array}{l}\mathrm{exp}\left(-\frac{1}{{x}^{2}}\right),\phantom{\rule{1em}{0ex}}x\ne 0\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{1em}{0ex}}x=0,\end{array}\right.}\end{array}$$

has an admissible zero of infinite multiplicity at *x* = 0 (see Remark 4 above).

An example of an isolated zero which is *not* admissible is given by the *C*^{∞}-function

$$\begin{array}{}{\displaystyle {f}_{3}(x):={f}_{2}(x)(\mathrm{sin}(\frac{1}{{x}^{3}})+2),}\end{array}$$

which vanishes (together with all derivatives) in 0 but the corresponding limits (2.1) do not exist.

#### Definition 2.3

A function *f* : [*a*, *b*] → ℝ belongs to 𝓐^{k}([*a*, *b*]), *k* ∈ ℕ, if the following holds:

*f* ∈ *C*^{0}([*a*, *b*]).

*f* has only admissible (and therefore finitely many) zeros *x*_{1} < … < *x*_{n} and *f*|_{(xi,xi+1)} (*i* = 1,…,*n*−1), *f*|_{(a,x1)} and *f*|_{(xn,b)} are of class *C*^{k+1}.

There exists a partition *a* = *y*_{1} < *y*_{2} < … < *y*_{m} = *b* such that *f*|_{(yi,yi+1)} is of class *C*^{k+2} for all *i* = 1,…,*m*−1.

If *f* ∈ 𝓐^{0}([*a*, *b*]), *f* will be called *admissible*.

#### Theorem 2.4

*Let f* ∈ 𝓐^{0}([*a*, *b*]). *The number of zeros n*(*f*) *of f in* [*a*, *b*] *is given by*

$$\begin{array}{}{\displaystyle n(f)=\underset{a}{\overset{b}{\int}}h\left(\frac{{f}^{\prime}(x)}{f(x)}\right)\frac{{f}^{\prime}(x{)}^{2}-f(x){f}^{\u2033}(x)}{f(x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+\underset{x\searrow b}{lim}\mathrm{H}\left(\frac{{f}^{\prime}(x)}{f(x)}\right)-\underset{x\nearrow a}{lim}\mathrm{H}\left(\frac{{f}^{\prime}(x)}{f(x)}\right)}\end{array}$$

*and the number of zeros n̊*(*f*) *of f in* (*a*, *b*) *by*

$$\begin{array}{}{\displaystyle \stackrel{\u02da}{n}(f)=\underset{a}{\overset{b}{\int}}h\left(\frac{{f}^{\prime}(x)}{f(x)}\right)\frac{{f}^{\prime}(x{)}^{2}-f(x){f}^{\u2033}(x)}{f(x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+\underset{x\nearrow b}{lim}\mathrm{H}\left(\frac{{f}^{\prime}(x)}{f(x)}\right)-\underset{x\searrow a}{lim}\mathrm{H}\left(\frac{{f}^{\prime}(x)}{f(x)}\right).}\end{array}$$

#### Proof

Consider first the case, where *f*(*a*),*f*(*b*) ≠ 0. Then the zeros of *f* are given by *a* < *x*_{1} < *x*_{2} < … < *x*_{n(f)} < *b*. The integrand of

$$\begin{array}{}{\displaystyle \underset{a}{\overset{b}{\int}}h\left(\frac{{f}^{\prime}(x)}{f(x)}\right)\frac{{f}^{\prime}(x{)}^{2}-f(x){f}^{\u2033}(x)}{f(x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=:\underset{a}{\overset{b}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x}\end{array}$$

is a priori undefined whenever *f* vanishes or whenever *f*″ is undefined. We decompose the integral and compute the resulting improper integrals using unilateral limits. Since *f* is admissible, we have

$$\begin{array}{}{\displaystyle \underset{{x}_{j}}{\overset{{x}_{j+1}}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\underset{x\searrow {x}_{j}}{lim}{\mathrm{H}}_{x}-\underset{x\nearrow {x}_{j+1}}{lim}{\mathrm{H}}_{x}=1}\end{array}$$

for all *j* = 1,…, *n*(*f*)−1, where H_{x}: = H(*f*′(*x*)/*f*(*x*)). Integrating over a neighborhood of a point *y* where *f*″ is undefined does not introduce further boundary terms since lim_{x ↘ y} H_{x} − lim_{x↗ y}H_{x} = 0. Hence

$$\begin{array}{}{\displaystyle \begin{array}{}\underset{a}{\overset{b}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\underset{a}{\overset{{x}_{1}}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+\sum _{j=1}^{n(f)-1}\underset{{x}_{j}}{\overset{{x}_{j+1}}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+\underset{{x}_{n(f)}}{\overset{b}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}={\mathrm{H}}_{a}-\underset{x\nearrow {x}_{1}}{lim}{\mathrm{H}}_{x}+(n(f)-1)+\underset{x\searrow {x}_{n(f)}}{lim}{\mathrm{H}}_{x}-{\mathrm{H}}_{b}\end{array}}\end{array}$$(2.5)

and therefore

$$\begin{array}{}{\displaystyle n(f)=\underset{a}{\overset{b}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+{\mathrm{H}}_{b}-{\mathrm{H}}_{a}.}\end{array}$$(2.6)

The computation above suggests that *n*(*f*) > 1 but one can check that formula (2.6) holds true for *n*(*f*) = 1 and *n*(*f*) = 0 as well.

If *f* has zeros in *a* and *b* and therefore *x*_{1} = *a*, *x*_{n}(*f*) = *b*, computation (2.5) gives

$$\begin{array}{}{\displaystyle n(f)=\underset{a}{\overset{b}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+1.}\end{array}$$(2.7)

According to (2.3),
$\begin{array}{}\underset{x\searrow b}{lim}{\mathrm{H}}_{x}-\underset{x\nearrow a}{lim}{\mathrm{H}}_{x}=1\end{array}$
H_{x} = 1 and (2.7) becomes

$$\begin{array}{}{\displaystyle n(f)=\underset{a}{\overset{b}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+\underset{x\searrow b}{lim}{\mathrm{H}}_{x}-\underset{x\nearrow a}{lim}{\mathrm{H}}_{x}}\end{array}$$

and hence (2.8) counts the zeros of *f* in [*a*, *b*] since it reduces to (2.6) if *f*(*a*),*f*(*b*) ≠ 0 and one can check that the remaining cases *f*(*a*) = 0≠ *f*(*b*) and *f*(*a*) ≠ 0 = *f*(*b*) are also covered. Let now

$$\begin{array}{}{\displaystyle \stackrel{\u02da}{n}(f)=\underset{a}{\overset{b}{\int}}\mathrm{I}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+\underset{x\nearrow b}{lim}{\mathrm{H}}_{x}-\underset{x\searrow a}{lim}{\mathrm{H}}_{x}.}\end{array}$$(2.8)

Since

$$\begin{array}{}{\displaystyle \mathrm{n!}(f)-\stackrel{\u02da}{n}(f)=\underset{x\searrow b}{lim}{\mathrm{H}}_{x}-\underset{x\nearrow a}{lim}{\mathrm{H}}_{x}-\left(\underset{x\nearrow b}{lim}{\mathrm{H}}_{x}-\underset{x\searrow a}{lim}{\mathrm{H}}_{x}\right)=}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\left\{\begin{array}{l}0,\phantom{\rule{1em}{0ex}}\text{if\hspace{0.17em}}f(a),f(b)\ne 0\\ 1,\phantom{\rule{1em}{0ex}}\text{if either\hspace{0.17em}}f(a)=0\text{\hspace{0.17em}or\hspace{0.17em}}f(b)=0\\ 2,\phantom{\rule{1em}{0ex}}\text{if\hspace{0.17em}}f(a)=f(b)=0\end{array}\right.}\end{array}$$

we conclude that *n̊*(*f*) counts the zeros of *f* in (*a*, *b*).□

Observe, that for a *C*^{2} function *f* with only zeros of multiplicity one, the integrand in (2.9) is continuous provided *h* is continuous. This remains true for zeros of higher multiplicity in the following way:

#### Proposition 2.5

*Let h*:ℝ → ℝ be continuous and *h*(*x*)
$\begin{array}{}\sim \frac{C}{{x}^{2}}for|x|\to \mathrm{\infty}.\end{array}$
*Then*, *the integrand in Theorem 2.4*

$$\begin{array}{}{\displaystyle \mathrm{I}:=h\left(\frac{{f}^{\prime}}{f}\right)\frac{{f}^{\prime 2}-f{f}^{\u2033}}{{f}^{2}}}\end{array}$$

*is continuous if f* ∈ *C*^{n}([*a*, *b*]), *n* ≥ 2, *has only zeros of multiplicity* ≤ *n*.

#### Proof

It suffices to show that I is continuous in 0 if *x* = 0 is a zero of *f* of multiplicity *n*. Then, by Taylor expansion, we have

$$\begin{array}{}{\displaystyle f(x)\text{\hspace{0.17em}}=\left(\frac{{f}^{(n)}(0)}{n!}+{r}_{0}(x)\right){x}^{n}}\\ {\displaystyle {f}^{\prime}(x)=\left(\frac{{f}^{(n)}(0)}{(n-1)!}+{r}_{1}(x)\right){x}^{n-1}}\\ {\displaystyle {f}^{\u2033}(x)=\left(\frac{{f}^{(n)}(0)}{(n-2)!}+{r}_{2}(x)\right){x}^{n-2}}\end{array}$$

where *r*_{i} are continuous functions with lim_{x → 0}*r*_{i}(*x*) = 0. Using these expressions in I, we get

$$\begin{array}{}{\displaystyle \mathrm{I}(x)=h\left(\frac{{s}_{1}(x)}{x}\right)\frac{{s}_{2}(x)}{{x}^{2}}}\end{array}$$

for continuous functions *s*_{i} with lim_{x → 0}*s*_{i}(*x*) = *n*. Thus

$$\begin{array}{}{\displaystyle \mathrm{I}(x)\sim \frac{C{x}^{2}}{{s}_{1}^{2}(x)}\frac{{s}_{2}(x)}{{x}^{2}}\to \frac{C}{n}}\end{array}$$

for *x* → 0.

If we only assume that *h*(*x*) = O(1/*x*^{2}) for |*x*| → ∞ in the previous proposition, the proof shows that then I is at least bounded.□

As a corollary of Proposition 2.5 we obtain that if *h* is continuous and *h*(*x*)
$\begin{array}{}h(x)\sim \frac{C}{{x}^{2}},\end{array}$
then I is in *C*^{0} provided *f* is analytic. Nontheless, the function *f* may behave in the neighborhood of a zero in such a pathological way, that I becomes unbounded (see Example 2.7.3). This is why, in general, the integrals in Theorem 2.4 have to be interpreted as improper integrals. This means that the concrete computation requires the zeros of *f* to be known a priori in order to evaluate the improper integrals. It is therefore of practical importance to formulate conditions (see Propositions 2.8 and 2.10) with additional assumptions which guarantee that I is in *L*^{1}: To this end we will slightly sharpen the admissibility condition for a function and impose some conditions on the behaviour of the zeros of *f*″ in neighborhoods of the zeros of *f*. Furthermore we will require *h* to have at least quadratic decay at infinity.

The proof of Proposition 2.5 for the case *C* = 1 indicates, how we can generalize the notion of multiplicity of zeros in a natural manner:

#### Definition 2.6

The multiplicity *μ*_{f}(*x*_{0}) of a zero *x*_{0} of *f* ∈ 𝓐^{0} is defined to be

$$\begin{array}{}{\displaystyle {\mu}_{f}({x}_{0})=\underset{x\to {x}_{0}}{lim}\frac{{f}^{\prime}(x{)}^{2}}{{f}^{\prime}(x{)}^{2}-f(x){f}^{\u2033}(x)}.}\end{array}$$

Since the zeros of functions in 𝓐^{0} are admissible, it follows that *μ*_{f}(*x*_{0}) ≥ qslant 0 whenever it exists, however, it can take values in [0,∞] (see Example 2.7.3 and 2.7.4 below). This definition of the multiplicity of a zero will be useful for a variant of Theorem 2.4 that takes the multiplicities of the zeros into account.

#### Example 2.7

A function *f* ∈ *C*^{n}, *n* ≥ 2 with 0 = *f*(*x*_{0}) = *f*′(*x*_{0}) = … = *f*^{(n−1)}(*x*_{0}) ≠ *f*^{(n)}(*x*_{0}) has a zero of multiplicity *n* in *x*_{0}: the Definition 2.6 is compatible with the usual notion of multiplicity.

The function *f*(*x*) = |*x*|^{r}, *r* > 0 has a zero of multiplicity *r* in *x* = 0.

The function

$$\begin{array}{}{\displaystyle f(x)=\left\{\begin{array}{l}\frac{1}{\mathrm{ln}|x|},x\ne 0\\ \phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\phantom{\rule{thinmathspace}{0ex}},x=0\end{array}\right.}\end{array}$$

has a zero of multiplicity 0 in *x* = 0.

The function *f*_{2} in Example 2.2.2 has a zero in *x* = 0 with *μ*_{f2}(0) = ∞.

#### Proposition 2.8

*Let h*:ℝ → ℝ *be a piecewise continuous function such that h*(*x*) = *O*(1/*x*^{2}) *for* |*x*| → ∞ *and let f* ∈ 𝓐^{0}([*a*, *b*]) ∩ *W*^{2,1}(*a*, *b*) *have only zeros of positive multiplicity in the sense of Definition 2.6. Furthermore we assume that for each zero x*_{0} *we have a neighborhood U such that either f*″(*x*)≡ 0 *on U*∖{*x*_{0}} *or*

$$\begin{array}{}{\displaystyle \sum _{k=1}^{\mathrm{\infty}}|{z}_{k}-{x}_{0}|<\mathrm{\infty},}\end{array}$$

*where z*_{1},*z*_{2},… *denote the countably many zeros of f*″ in *U. Then*

$$\begin{array}{}{\displaystyle \mathrm{I}:=h\left(\frac{{f}^{\prime}}{f}\right)\frac{{f}^{\prime 2}-f{f}^{\u2033}}{{f}^{2}}\in {L}^{1}(a,b).}\end{array}$$

#### Proof

Choose neighborhoods *U*_{1},…, *U*_{n} of the *n* zeros of *f*, which do not (with the possible exception of the respective zero itself) contain singular points of *f*″ or zeros of *f*′ and let

$$\begin{array}{}{\displaystyle \mathrm{U}=\bigcup _{i=1}^{n}{U}_{i}.}\end{array}$$

Since |*f*| ≥ *η* for some *η* > 0 on the complement U^{c} and *W*^{2,1}(*a*, *b*)↪ *C*^{1}([*a*, *b*]) we can estimate

$$\begin{array}{}{\displaystyle \underset{{\mathrm{U}}^{c}}{\int}|\mathrm{I}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x={\eta}^{-2}\parallel h{\parallel}_{{L}^{\mathrm{\infty}}(\mathbb{R})}(\parallel {f}^{\prime 2}{\parallel}_{{C}^{0}([a,b])}|b-a|+\parallel f{\parallel}_{{C}^{0}([a,b])}\parallel {f}^{\u2033}{\parallel}_{{L}^{1}(a,b)})<\mathrm{\infty}.}\end{array}$$

Consider now wlog the neighborhood *U*_{i} of the zero *x*_{i} = 0 and assume *U*_{i} = (−*ε*,*ε*) for some *ε* > 0. We need to show that I|_{(−ε,ε)} ∈ *L*^{1}. Since *h*(*x*) = O(1/*x*^{2}) for |*x*| → ∞, there exists a constant *C* > 0 such that

$$\begin{array}{}{\displaystyle |\mathrm{I}(x)|\u2a7dC\left(1+\left|\frac{f(x){f}^{\u2033}(x)}{{f}^{\prime}(x{)}^{2}}\right|\right).}\end{array}$$(2.10)

Note that *f* *f*″/*f*′^{2} ∈ *L*^{1}(−*ε*,*ε*) if and only if N ∈ BV(−*ε*,*ε*), where N(*x*) = *x*−*f*(*x*)/*f*′(*x*) denotes the *Newton-Operator* of *f* and BV(−*ε*,*ε*) denotes the space of functions *g*:(−*ε*,*ε*) → ℝ of bounded variation. It follows from the admissibility of the zero that N:(−*ε*,*ε*)∖{0} → ℝ can be continuously extended to N(0) = 0 and it holds that

$$\begin{array}{}{\displaystyle {\mathrm{N}}^{\prime}(x)=\frac{f(x){f}^{\u2033}(x)}{{f}^{\prime}(x{)}^{2}},}\end{array}$$

for *x* ≠ 0. Let *μ* > 0 denote the multiplicity of the zero according to Definition 2.6. It holds that

$$\begin{array}{}{\displaystyle \underset{x\to 0}{lim}{\mathrm{N}}^{\prime}(x)=\left\{\begin{array}{l}\frac{\mu -1}{\mu},\mu <\mathrm{\infty}\\ \phantom{\rule{2em}{0ex}}\text{\hspace{0.17em}}1\phantom{\rule{thinmathspace}{0ex}},\mu =\mathrm{\infty}.\end{array}\right.}\end{array}$$

According to the mean value theorem we have N(*x*)/*x* = N′(*ξ*) for some *ξ* between 0 and *x* and deduce that N ∈ *C*^{1}(−*ε*,*ε*). The Taylor expansion of N around *x* = 0 is given by

$$\begin{array}{}{\displaystyle \mathrm{N}(x)=\left\{\begin{array}{l}\frac{\mu -1}{\mu}x+o(x),\text{\hspace{0.17em}}\mu <\mathrm{\infty}\\ \phantom{\rule{2em}{0ex}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+o(x),\text{\hspace{0.17em}}\mu =\mathrm{\infty}.\end{array}\right.}\end{array}$$

In any case there exists a constant *K* > 0 such that

$$\begin{array}{}{\displaystyle |\mathrm{N}(x)|\le K|x|,\phantom{\rule{1em}{0ex}}|x|<\epsilon .}\end{array}$$(2.11)

We will now show that N ∈ BV([0,*ε*)), the argument on (−*ε*,0] being similar. We start by noticing that N is absolutely continuous on [*δ*,*ε*) for every 0 < *δ* < *ε* since *x*, *f*(*x*) and *f*′(*x*) are absolutely continuous and *f*′(*x*) ≠ 0 on [*δ*,*ε*). In particular, N ∈ BV([*δ*,*ε*)) for every 0 < *δ* < *ε*.

We will now distinguish two cases: If *f*″≡ 0 on (0,*ε*), then N ≡ 0 and we are done. In the remaining case we first consider the case when the set of zeros of *f*″ in (0,*ε*) is empty: Then N is monotone on [0,*ε*) and hence N ∈ BV([0,*ε*)). Otherwise the zeros of *f*″ in [0,*ε*) are given by *z*_{1} > *z*_{2} > … and we may set *δ*: = *z*_{1}. According to (2.11) and since the zeros of *f*″ are precisely the zeros of N′ we can estimate the total variation of N on (*z*_{k+1},*z*_{k}) by

$$\begin{array}{}{\displaystyle \underset{{z}_{k+1}}{\overset{{z}_{k}}{\int}}|{\mathrm{N}}^{\prime}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\u2a7d2K{z}_{k}.}\end{array}$$

The total variation of N on [0,*ε*) is bounded by

$$\begin{array}{}{\displaystyle \sum _{k=1}^{\mathrm{\infty}}\underset{{z}_{k+1}}{\overset{{z}_{k}}{\int}}|{\mathrm{N}}^{\prime}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x+\underset{\delta}{\overset{\epsilon}{\int}}|{\mathrm{N}}^{\prime}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\u2a7d2K\sum _{k=1}^{\mathrm{\infty}}{z}_{k}+\underset{\delta}{\overset{\epsilon}{\int}}|{\mathrm{N}}^{\prime}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x,}\end{array}$$

where the series converges by assumption and the integral is finite since N ∈ BV([*δ*,*ε*)). We conclude that N ∈ BV([0,*ε*)), which finishes the proof.□

#### Example 2.9

Let

$$\begin{array}{}{\displaystyle k(x)=\left\{\begin{array}{l}{x}^{3}+\left(\sqrt{|x{|}^{7}}-{x}^{3}\right)\mathrm{cos}\left(\pi {\mathrm{log}}_{2}|x|\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}x\ne 0\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}x\ne 0.\end{array}\right.}\end{array}$$

Then *f*(*x*) =
$\begin{array}{}{\int}_{0}^{x}\end{array}$
*k*(*t*) d*t* is of class *C*^{3} and has an admissible zero in *x* = 0 but *f*(*x*)/(*x**f*′(*x*)) is unbounded near 0.

#### Proposition 2.10

*Let h*:ℝ → ℝ *be a piecewise continuous function such that h*(*x*) = O(1/*x*^{2}) *for* |*x*| → ∞ *and let f* ∈ 𝓐^{0}([*a*, *b*])∩ *W*^{2,1}(*a*, *b*) *be such that that for every zero x*_{0} *of f there exists a relatively open neighborhood U* ⊂ [*a*, *b*] *such that*

$$\begin{array}{}{\displaystyle 0<\left|\frac{f(x)}{(x-{x}_{0}){f}^{\prime}(x)}\right|<\stackrel{~}{K}}\end{array}$$(2.12)

*on U* ∖{*x*_{0}} *and such that either f*″≡ 0 *on U*∖{*x*_{0}}, or

$$\begin{array}{}{\displaystyle \sum _{k=1}^{\mathrm{\infty}}|{z}_{k}-{x}_{0}|<\mathrm{\infty},}\end{array}$$

*where z*_{1},*z*_{2},… *denote the countably many zeros of f*″ *in U*∖{*x*_{0}}. *Then*

$$\begin{array}{}{\displaystyle \mathrm{I}:=h\left(\frac{{f}^{\prime}}{f}\right)\frac{{f}^{\prime 2}-f{f}^{\u2033}}{{f}^{2}}\in {L}^{1}(a,b).}\end{array}$$

#### Proof

Choose neighborhoods *U*_{1},…, *U*_{n} of the *n* zeros of *f*, which do not (with the possible exception of the respective zero itself) contain singular points of *f*″ or zeros of *f*′ such that (2.12) holds on each punctured neighborhood. As in the proof of Proposition 2.8 we obtain ∥I∥_{L1(Uc)} < ∞, where U = *U*_{1}∪ … ∪ *U*_{n} and the estimate (2.10). Let wlog 0 be a zero of *f* and let (−*ε*,*ε*) be its respective neighborhood for some *ε* > 0. As in the proof of Proposition 2.8, we are done if we show that N ∈ BV([0,*ε*)). The condition 0 < |*f*(*x*)/(*x* *f*′(*x*))| < *K͠* on (−*ε*,*ε*)∖{0} implies that

$$\begin{array}{}{\displaystyle 0<\left|\frac{f(x)}{{f}^{\prime}(x)}\right|<\stackrel{~}{K}|x|,}\end{array}$$(2.13)

from which we conclude that *N* extends continuously to [0,*ε*) (where N(0) = 0) and

$$\begin{array}{}{\displaystyle |\mathrm{N}(x)|\u2a7d(\stackrel{~}{K}+1)\phantom{\rule{thinmathspace}{0ex}}x,\phantom{\rule{1em}{0ex}}x\in [0,\epsilon ).}\end{array}$$(2.14)

This is just estimate (2.11) with *K* = *K͠* + 1. The rest of the proof is exactly the same as the one of Proposition 2.8.□

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