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Let $I$ be an interval and $f:I\to \mathbb{R}$ function that has jump discontinuity at some interior point $x_0\in I$, i.e limits $f(x_0^+)$ and $f(x_0^-)$ exist, but are not equal. Prove that $f$ does not have a primitive on $I.$

Comment. Proving this answers another quenstion of mine, if the Intermediate Value Property (necessary condition for existence of primitive) is sufficient for the existence of primitive. Then the answer would be no: take $f(x)=x$ on $[0,1]$ and $f(x)=3-x$ on $(1,2]$, which enjoys the IMP but has jump discontinuity at $x=1$ and therefore no primitive exists (of course that may have been done from the scratch).

Getting back to the question, by contradiction, if a primitive of $f$ existed, then $f$ would enjoy the IVP. But how does this contradict the jump discontinuity of $f$?

Thanks in advance.

Nikolaos Skout
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    Your example does not seem to satisfy the IMP. $f(1/2)=1/2$ and $f(3/2)=3/2$, but there is no $x$ between $1/2$ and $3/2$ such that $f(x)=5/4$. – mickep Nov 21 '19 at 12:01
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    This post, says, that every derivative satisfies the intermediate value property: https://math.stackexchange.com/questions/1776315/intermediate-value-for-derivative-apostol-text This is also related: https://math.stackexchange.com/questions/327243/is-every-function-with-the-intermediate-value-property-a-derivative?r=SearchResults – humanStampedist Nov 21 '19 at 12:46

1 Answers1

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If $f$ has a primitive $F$, by definition $f$ satisfies $F'=f$. Now, by Darboux theorem $f$ satisfies the IVP.

A function with a jump discontinuity cannot satisfy the IVP:

  • First proof: analytical

    let $\varepsilon=\frac{|f(x_0^-)-f(x_0^+)|}{3}$. By limit definition, there exists $\delta_1,\delta_2$ such that $x_0-\delta_1<x<x_0\rightarrow |f(x_0^-)-f(x)|<\varepsilon$ and $x_0<x<\delta_2+x_0\rightarrow |f(x_0^+)-f(x)|<\varepsilon$. Setting $\delta=\min(\delta_1,\delta_2)$ we obtain that

$$f\big((x_0-\delta,x_0+\delta)\big)\subset \big(f(x_0^-)-\varepsilon,f(x_0^-)+\varepsilon\big)\cup \big(f(x_0^+)-\varepsilon,f(x_0^-)+\varepsilon\big)\cup {f(x_0)}$$

But this contradicts the IVP.

  • Second proof: topological

We claim that a function $h:\mathbb{R}\to \mathbb{R}$ satisfies the IVP only if it maps closed intervals to intervals (in $\mathbb{R}^*$). The fact that $h$ cannot have a jump discontinuity then trivially follows.

To prove our claim, let $h([a,b])$ be a set that is not an interval (and thus it is not path connected, since the only path connected sets in $\mathbb{R}*$ are the intervals). Then there is a point $y\in[h(a),h(b)]$ that is not in $h([a,b])$, and this clearly contradicts the IVP.

  • Note:On the existence of primitive

To answer your other question: the IVP is not enough to ensure the existence of a primitive. Actually, it is not enough to even ensure the non boundedness of the function (see, for example, the Base 13 function).

  • Note: A proof of Darboux Theorem

The case in which $f=F'=k'$ is trivial. Let $F$ be a differentiable mom constant function defined on a closed interval $[a,b]$ and let $f=F'$, and let $y$ be a number in $(f(a),f(b))$ (if instead $f(b)<f(a)$ the proof is no different). Now, let $h(x):=F(x)-xy$. This function is continuous and so, by Weierstrass theorem, it has a maximum and a minimum. They can't both be on the boundary of the interval, since this would imply the fact that $F$ is constant. Thus, at least one of those point is on the interior of the interval, and by Fermat theorem in this point (let us call it $\zeta$) we have $h'(\zeta)=0\rightarrow f(\zeta)-y=0\rightarrow f(\zeta)=y$.

Michael Hardy
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