converse of Darboux’s theorem (analysis) is not true

Darboux’ theorem says that, if $f\\colon\\mathbb{R}\\rightarrow\\mathbb{R}$ has an antiderivative, than $f$ has to satisfy the intermediate value property, namely, for any $a<b$ , for any number $C$ with $f(a)<C<f(b)$ or $f(b)<C<f(a)$ , there exists a $c\\in(a,b)$ such that $f(c)=C$ . With this theorem, we understand that if $f$ does not satisfy the intermediate value property, then no function $F$ satisfies $F^{\\prime}=f$ on $\\mathbb{R}$ .

Now, we will give an example to show that the converse is not true, i.e., a function that satisfies the intermediate value property might still have no antiderivative.

Let

f(x)=\\left\\{\\begin{array}[]{ccr}\\displaystyle\\frac{1}{x}\\cos(\\ln x)&\\mbox{if}&%x>0\\\\0&\\mbox{if}&x\\leq 0\\end{array}\\right..

First let us see that $f$ satisfies the intermediate value property. Let $a<b$ . If $0<a$ or $b\\leq 0$ , the property is satisfied, since $f$ is continuous on $(-\\infty,0]$ and $(0,\\infty)$ . If $a\\leq 0<b$ , we have $f(a)=0$ and $f(b)=(1/b)\\cos(\\ln b)$ . Let $C$ be between $f(a)$ and $(b)$ . Let $a_{0}=\\exp(-2\\pi k_{0}+\\pi)$ for some $k_{0}$ large enough such that $a_{0}<b$ . Then $f(a_{0})=0=f(a)$ , and since $f$ is continuous on $(a_{0},b)$ , we must have a $c\\in(a_{0},b)$ with $f(c)=C$ .

Assume, for a contradiction that there exists a differentiable function $F$ such that $F^{\\prime}(x)=f(x)$ on $\\mathbb{R}$ . Then consider the function $G(x)=\\sin(\\ln x)$ which is defined on $(0,\\infty)$ .We have $G^{\\prime}(x)=f(x)$ on $(0,\\infty)$ , and since it is a an open connected set, we must have $F(x)=G(x)+c$ on $(0,\\infty)$ for some $c\\in\\mathbb{R}$ . But then, we have

\\displaystyle\\limsup_{x\\rightarrow 0^{+}}F(x)

\\displaystyle=\\limsup_{x\\rightarrow 0^{+}}G(x)+c=1+c

and

\\displaystyle\\liminf_{x\\rightarrow 0^{+}}F(x)

\\displaystyle=\\liminf_{x\\rightarrow 0^{+}}G(x)+c=-1+c

which contradicts the differentiability of $F$ at $0$ .