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Since differentiation is linear, we therefore have that if $f, g: I\to \mathbb{R}$ is a derivative (where $I\subset \mathbb{R}$ is an interval), then so does their linear combination. What if we consider their multiplication and composition?

Due to the forms of the product rule of differentiation of product function and chain rule of differentiation of composition, I highly doubt their product or composition necessarily is still a derivative, but I cannot construct counterexamples.

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  • $\begingroup$ Hint for a complete answer for closed-form antiderivatives: J. Liouville's theorem on integration in finite terms solves the problem for antiderivatives which are elementary functions. One can apply Liouville's method to generalize this to arbitrary classes of closed-form functions. One has to apply Liouville's theorem and method to the product and composition of functions. The case of infinite terms (infinite series, products, fractions, ...) still remains. $\endgroup$ – IV_ Dec 30 '16 at 23:16
  • $\begingroup$ To clarify, does " $f$ has an antiderivative" mean that $\int f(x)\;dx$ exists, or that it can be written in closed form? $\endgroup$ – MPW Jan 3 '17 at 21:40
  • $\begingroup$ @MPW Those of us who write on such subjects prefer the phrase "$f$ is a derivative'' rather than "$f$ has an antiderivative." Both mean that there exists a function $F$ with $F'(x)=f(x)$ at every point $x$ in some interval (that really should be specified). A "closed form" or formula is not an issue in such discussions. Usually words like "antiderivative" or "indefinite integral" [in this sense] and notation like $\int f(x)\,dx$ are found only in calculus classes, not real analysis research. $\endgroup$ – B. S. Thomson Jan 6 '17 at 17:38
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Let me address just one of your problems.

Problem. Suppose that $f$ and $g$ are both derivatives. Under what conditions can we assert that the product $fg$ is also a derivative?

The short answer is that this is not true in general. In fact even if we assume that $f$ is continuous and $g$ is a derivative the product need not be a derivative. However if we strengthen that to assuming that $f$ is not merely continuous but also of bounded variation, then indeed the product with any derivative would be a derivative.

This is an interesting problem and leads to interesting ideas.

For references to the statements here and an in depth look at the problem here are some references:

Bruckner, A. M.; Mařík, J.; Weil, C. E. Some aspects of products of derivatives. Amer. Math. Monthly 99 (1992), no. 2, 134–145.

Fleissner, Richard J. Distant bounded variation and products of derivatives . Fund. Math. 94 (1977), no. 1, 1–11.

Fleissner, Richard J. On the product of derivatives. Fund. Math. 88 (1975), no. 2, 173–178.

Fleissner, Richard J. Multiplication and the fundamental theorem of calculus—a survey. Real Anal. Exchange 2 (1976/77), no. 1, 7–34.

Foran, James On the product of derivatives, Fund. Math. 80 (1973), no. 3, 293–294.

I will edit in some links when I find them. Foran and Fleissner were close childhood friends who ended up pursuing their PhD at the same time in Milwaukee. Fleissner died in an automobile accident in 1983.

NOTE ADDED. Elementary students are not going to want to pursue this topic to quite this depth. But here is an exercise aimed at this level that they might find entertaining.

Exercise. Consider the function $$f(x)=\begin{cases} \cos \frac1x, & x\not=0 \\ 0 &x=0 \end{cases} $$ Show that the function $f$ is a derivative but that its square $f^2$ is not.

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    $\begingroup$ The second problem is less interesting and I haven't mentioned it. But in order for $\phi(f(x))$ to be a derivative whenever $f$ is a derivative it is only possible if $\phi$ is a linear function (as one likely would guess). $\endgroup$ – B. S. Thomson Dec 12 '15 at 18:13
  • $\begingroup$ I noticed your comment only after I placed the bounty! If you add a proof of what you said in your comment to your answer, I will gladly give you the +250. $\endgroup$ – MathematicsStudent1122 Dec 28 '16 at 22:42
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    $\begingroup$ Here are a couple of papers of A M Bruckner where such problems are discussed: A M Monthly, 85, 1978, 554-562 and Quart J Math Oxford 29, 1978, 1-10. Andy gives Choquet credit for the observation, but Choquet did not supply a proof. Don't need a bounty, but would accept a bottle of wine. $\endgroup$ – B. S. Thomson Dec 30 '16 at 21:17
  • $\begingroup$ @MathematicsStudent1122 Next time you offer a bounty, please make it clear what is required in the answer to be awarded the bounty (or a part thereof). I'm not an expert in this area like our contributor B.S.Thomson, but I spent some time on the problem, prodded on by your bounty, and thought I came up with something of interest, leading to a nice example addressing the composition question. Bounty awarded: $0.$ Weird. $\endgroup$ – zhw. Jan 9 '17 at 23:57
  • $\begingroup$ @zhw. That was really a mistake. I actually forgot about this question, and it seems I must have missed the bounty-expiring notification. Sorry. I've +1'd your answer. $\endgroup$ – MathematicsStudent1122 Jan 10 '17 at 0:44
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Let $X$ be the set of functions from $\mathbb R$ to $\mathbb R$ that have antiderivatives on all of $\mathbb R.$

Claim: There exists $f\in X$ such that $f^n\notin X$ for $n=2,3,\cdots.$ (Here $f^n$ is the $n$th power of $f,$ not the $n$th iterate.) This addresses both the product question and the composition question. If the claim is true, then $f\in X$ but the product $f\cdot f \notin X.$ Also, with $g_n(x) = x^n, n = 2,3, \dots,$ we see $g_n\circ f\notin X$ even though the $g_n$'s are very nice and $f\in X.$

The following lemma will be helpful:

Lemma: Suppose $f:\mathbb R\to \mathbb R,$ with $f=0$ on $(-\infty,0]$ and $f$ continuous on $(0,\infty).$ Then $f\in X$ iff $\int_0^x f$ converges for all $x>0$ and

$$\lim_{x\to 0^+}\frac{1}{x} \int_0^x f = 0.$$

Here the convergence of $\int_0^x f$ is in the improper integral sense, meaning $\int_0^x f = \lim_{a\to 0^+}\int_a^xf$ assuming the limit is finite.

Proof: $\implies:$ Let $f\in X,$ with $F'=f$ on $\mathbb R.$ Let $x>0.$ For any $a\in (0,x),$ $F\in C^1[a,x].$ Thus

$$\tag 1 F(x)-F(a) = \int_a^x f.$$

As $a\to 0^+,$ the left side of $(1)$ $\to F(x) - F(0)$ by the continuity of $F.$ It follows that the right side has a finite limit, hence $\int_0^x f$ converges. Note that we have shown

$$F(x)-F(0) = \int_0^x f.$$

Now $F'(0) = f(0)= 0.$ It follows that

$$0 = \lim_{x\to 0+}\frac{F(x)-F(0)}{x-0}= \lim_{x\to 0+} \frac{\int_0^x f}{x}$$

as desired.

$\impliedby:$ Simply define $F(x) = \int_0^x f.$ By the continuity of $f$ on $\mathbb R\setminus \{0\},$ we have $F'(x) = f(x)$ for $x\ne 0.$ At $0,$ we easily see the derivative from the left of $F$ at $0$ is $0.$ From the right we have

$$\lim_{x\to 0+}\frac{F(x)-F(0)}{x-0}= \lim_{x\to 0+} \frac{\int_0^x f}{x} = 0.$$

This shows $f\in X$ and the proof of the lemma is complete.

Proof of the claim: We build a slightly crazy $f\in X.$ For $n=2,3,\dots$ let $a_n = 1/n, b_n = 1/n + 1/2^n.$ Then the intervals $[a_n,b_n]$ are pairwise disjoint.

On each $[a_n,b_n]$ let $f$ be an isoceles triangular spike from the end points of height $2^n/n^3.$ Define $f =0$ everywhere else. Then $f$ satisfies the hypotheses of the lemma. Now

$$\int_0^1 f = \sum_{n=2}^{\infty} \frac{1}{2}\cdot \frac{1}{2^{n+1}}\cdot \frac{2^n}{n^3} = \sum_{n=2}^{\infty} \frac{1}{4n^3} <\infty.$$

To show $(1/x)\int_0^x f \to 0,$ suppose $x\in [a_n,a_{n-1}].$ Then

$$\tag 2 \frac{1}{x}\int_0^x f \le \frac{1}{a_n}\int_0^{a_{n-1}}f = n \sum_{k=n}^{\infty}\frac{1}{4n^3}.$$

The last sum is $O(1/n^2).$ So $(2)$ is no more than $nO(1/n^2) = O(1/n) \to 0.$ Thus the left side of $(2)\to 0$ as $x\to 0^+.$ The lemma now shows $f\in X.$

The easy part of the claim is showing higher powers of $f$ are not in $X.$ By the lemma, all we need to show is that $\int_0^1 f^2 = \infty.$ It will then follow that $\int_0^1 f^n = \infty, n=3,4,\dots $

Here's what I get (but the reader should check):

$$\int_{a_n}^{b_n} f^2 = \frac{2^n}{3n^6} .$$

Thus $\int_0^1 f^2 =\infty,$ by a mile. The proof of the claim is complete.

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  • $\begingroup$ I believe the following generalization is true: Let $p>1.$ Suppose $g$ is continuous on $\mathbb R,$ with $|g(x)|>x^p$ for large $x>0,$ then $g\circ f \notin X.$ (Here $f\in X$ is the function constructed in my answer.) Example: if $P$ is any nonlinear polynomial, then $P\circ f \notin X.$ $\endgroup$ – zhw. Dec 31 '16 at 19:36
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From the product rule of differentiation or from the rule of partial integration, one obtains the following less known product rules:

\begin{equation}\int f(x)g(x)dx=f(x)G(x)-\int f'(x)G(x)dx\hspace{1.5cm}[Borman], [Goldstein], [Kowalk]\end{equation} \begin{equation}\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx\hspace{3.9cm}[Mercer], [Reinsch]\end{equation}

$F(z)=\int f(z)dz$; $G(z)=\int g(z)dz$; $\gamma$: compositional inverse of the function $g$

From the chain rule of differentiation, the inverse rule of differentiation and the rule of partial integration, one obtains the following less known chain rules: \begin{equation}\int f(g(x))dx=\int f(t)\gamma'(t)dt;\ t=g(x)\end{equation} \begin{equation}\int f(g(x))dx=xf(g(x))-\int f'(t)\gamma(t)dt;\ t=g(x)\end{equation} \begin{equation}\int f(g(x))dx=\left(\frac{d}{dx}F(g(x))\right)\int\frac{1}{g'(x)}dx-\int \left(\frac{d^{2}}{dx^{2}}F(g(x))\right)\int\frac{1}{g'(x)}dx\ dx\end{equation} \begin{equation}\int f(g(x))dx=\frac{F(g(x))}{g'(x)}+\int F(g(x))\frac{g''(x)}{g'(x)^{2}}dx\end{equation}

This formulas show that integration rules for products and compositions of functions are possible only for some special kinds of component functions $f$ and $g$.

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