# Smooth sequence of functions converging pointwise to a smooth function and limit of derivatives

In the wikipedia page of uniform convergence, it says that given a sequence $$\{ f_n \}$$ of differentiable real functions (say, over the reals) with the property that it converges pointwise to some function $$f$$, the limit of $$\{f_{n}' \}$$ need not be equal to $$f'$$.

It then gives an example where $$\{ f_n \}$$ converges uniformly to a differentiable $$f$$, but $$\{f_n '\}$$ does not converge even pointwise.

My question is, what if we assume that each $$f_n$$ and their limit $$f$$ are, say, $$C^\infty$$, and $$\{ f_n '\}$$ converges pointwise to some $$C^\infty$$ function $$g$$ as well. Is it now enough to show that $$f' = g$$? Or do we further need to assume uniform convergence? Is there a classic counterexample to this question as well?

• Intersting question! The answer is yes if you know also that $|f_n'|\leq M$ for some constant independent of $n$. I'm not sure right now about the general case. Sep 14, 2021 at 21:27
• @Jose27 Oh right, that's like a special case of dominated convergence? Sep 14, 2021 at 21:30
• Yep, that's right. Of course, you can get away with a weaker condition like $|f_n'|\leq h$ for some integrable function $h$, but it's still not enough to give the general case unfortunately. Sep 14, 2021 at 21:33
• @Jose27 So this could mean it is likely there is a counterexample obtained from a sequence not satisfying the criteria for dominated convergence. Sep 14, 2021 at 21:35
• It also looks like this should be useful, then we can reduce to taking integrals over compact intervals: math.stackexchange.com/questions/3100416/… Sep 14, 2021 at 21:46

Let $$f_1,f_2,\ldots$$ be a sequence of $$C^\infty$$ functions from $${\mathbb R}$$ to $${\mathbb R}$$.

Let $$f,g$$ be two more $$C^\infty$$ functions from $${\mathbb R}$$ to $${\mathbb R}$$.

Assume, as $$i\to\infty$$, $$f_i\to f$$ pointwise.

Assume, as $$i\to\infty$$, $$f'_i\to g$$ pointwise.

We wish to show: $$f'=g$$.

Define $$G:{\mathbb R}\to{\mathbb R}$$ by: $$\forall x\in{\mathbb R}$$, $$G(x)=\int_0^x\,g$$. Then $$G$$ is $$C^\infty$$ and $$G'=g$$.

For all integers $$i>0$$, let $$\phi_i:=f_i-G$$; then $$\phi_i$$ is $$C^\infty$$.

Also, $$\forall$$integers $$i>0$$, we have: $$\phi'_i=f'_i-g$$.

Let $$\phi:=f-G$$; then $$\phi$$ is $$C^\infty$$. Also, we have: $$\phi'=f'-g$$.

Also, as $$i\to\infty$$, $$\phi_i\to\phi$$ pointwise. Also, $$\phi'_i\to0$$ pointwise.

It suffices to show: $$\phi'=0$$.

Given $$a\in{\mathbb R}$$, we wish to show that $$\phi'(a)=0$$. Assume $$\phi'(a)\ne0$$. Want: Contradiction.

Let $$\varepsilon:=|\phi'(a)|/2$$. Then $$|\phi'(a)|>\varepsilon>0$$.

By continuity of $$\phi'$$, choose $$\delta>0$$ s.t., on $$[a-\delta,a+\delta]$$, $$|\phi'|>\varepsilon$$.

Let $$K:=[a-\delta,a+\delta]$$. Then, on $$K$$, $$|\phi'|>\varepsilon$$.

For all integers $$n>0$$, let $$C_n:=\{x\in K~\hbox{s.t.}~\forall\hbox{integers }i\ge n,\,|\phi'_i(x)|\le\varepsilon\};$$ then $$C_n$$ is closed in $${\mathbb R}$$ and $$C_n\subseteq K$$.

Because $$\phi'_i\to0$$ pointwise, we get: $$C_1\cup C_2\cup C_3\cup\cdots=K$$.

By the Baire Category Theorem, choose an integer $$n>0$$ s.t.: the interior in $${\mathbb R}$$ of $$C_n$$ is nonempty.

Choose $$q$$ in the interior in $${\mathbb R}$$ of $$C_n$$. Choose $$\rho>0$$ s.t. $$[q-\rho,q+\rho]\subseteq C_n$$.

Let $$J:=[q-\rho,q+\rho]$$. Then $$J\subseteq C_n$$.

By definition of $$C_n$$, $$\forall$$integers $$i\ge n$$, we have: on $$C_n$$, $$|\phi'_i|\le\varepsilon$$.

Therefore, $$\forall$$integers $$i\ge n$$, we have: on $$J$$, $$|\phi'_i|\le\varepsilon$$.

Then, by the Mean Value Theorem, $$\forall$$integers $$i\ge n$$, $$\forall$$distinct $$a,b\in J$$, $$\left|\frac{\phi_i(b)-\phi_i(a)}{b-a}\right|\le\varepsilon.$$ Taking the limit, as $$i\to\infty$$, we get: $$\forall$$distinct $$a,b\in J$$, $$\left|\frac{\phi(b)-\phi(a)}{b-a}\right|\le\varepsilon.$$ So, since $$q\in J$$, we get: $$\forall b\in J\backslash\{q\}$$, $$\left|\frac{\phi(b)-\phi(q)}{b-q}\right|\le\varepsilon.$$ Letting $$b\to q$$, we get: $$|\phi'(q)|\le\varepsilon$$.

Recall: on $$K$$, $$|\phi'|>\varepsilon$$. So, since $$q\in J\subseteq C_n\subseteq K$$, we get $$|\phi'(q)|>\varepsilon$$.

• Sorry for the delay! Thanks for the excellent answer May 30, 2022 at 0:40
• After defining $C_n$, you wrote: "Since $\phi_i \to 0$ pointwise," when you meant "Since $\phi'_i \to 0$ pointwise," Jun 19, 2022 at 20:33
• I presume that the same argument works for multivariable cases as well? Jun 21, 2023 at 14:33
• I suppose one natural generalization would be to have a fixed $C^\infty$ vector field $X$ on a manifold $M$, and a sequence $f_1,f_2,\ldots:M\to{\mathbb R}$ of $C^\infty$ functions, along with two more $C^\infty$ functions $f:M\to{\mathbb R}$ and $g:M\to{\mathbb R}$. Assume $f_i\to f$ pointwise on $M$ as $i\to\infty$. Assume $Xf_i\to g$ pointwise on $M$. Then $Xf=g$ on $M$. Is this the kind of thing you have in mind? I do think this should be true. If it were to fail at a point $p\in M$, then, using an integral curve for $X$ through $p$, it seems to me that we'd contradict the main result. Jun 22, 2023 at 20:57

Thank you for pointing out a typo, Yuval. I corrected it.

Trying to minimize hypotheses, in the proof, I came up with:

Let $$f_1,f_2,f_3,\ldots:{\mathbb R}\to{\mathbb R}$$ be a sequence of functions all differentiable on $${\mathbb R}$$. Assume that the pointwise limit $$f$$ of $$f_1,f_2,f_3,\ldots$$ exists everywhere on $${\mathbb R}$$ and is continuously differentiable on $${\mathbb R}$$, and that the pointwise limit $$g$$ of $$f'_1,f'_2,f'_3,\ldots$$ exists everywhere on $${\mathbb R}$$ and is continuous on $${\mathbb R}$$. Then $$f'=g$$.

First, if we remove the hypothesis of continuity of $$g$$, we get a counterexample, as follows.

Define $$f_1,f_2,f_3,\ldots:{\mathbb R}\to{\mathbb R}$$ by: $$\forall$$integers $$k\ge1$$, $$\forall x\in{\mathbb R}$$, $$f_k(x)=xe^{-kx^2}$$.

Then the pointwise (or, even, uniform) limit $$f$$ of $$f_1,f_2,f_3,\ldots$$ is the constant function $$0$$ on $${\mathbb R}$$.

However, the pointwise limit $$g$$ of $$f'_1,f'_2,f'_3,\ldots$$ is equal to $$1$$ at $$0$$, and is equal to $$0$$ on $${\mathbb R}\backslash\{0\}$$.

Then $$f'=g$$ fails to be true at $$0$$.

Second, we might only assume that $$f_1,f_2,f_3,\ldots$$ converges pointwise to a continuous limit $$f$$, keeping the hypothesis that $$f'_1,f'_2,f'_3,\ldots$$ converges pointwise to a continuous limit $$g$$, and then hope to prove continuous differentiability everywhere on $${\mathbb R}$$ of $$f$$. If we could prove such a result, then the theorem above would give: $$f'=g$$.

However, we get a counterexample, as follows.

Let $$J:=[0,1]$$.

Let $$U_1:=(1/3,2/3)$$.

Let $$U_2:=(1/9,2/9)\cup(7/9,8/9)$$.

Let $$U_3:=(1/27,2/27)\cup(7/27,8/27)\cup(19/27,20/27)\cup(25/27,26/27)$$.

etc.

For any integer $$k\ge1$$, to go from $$U_k$$ to $$U_{k+1}$$, write $$J\backslash(U_1\cup\cdots\cup U_k)$$ as a finite union of $$2^k$$ compact intervals, and let $$U_{k+1}$$ be the union of the "open middle thirds" of each of those compact intervals.

Then $$U_1,U_2,U_3,\ldots$$ are pairwise-disjoint.

The Cantor set is $$C:=J\backslash(U_1\cup U_2\cup U_3\cup\cdots)$$. Let $$\phi$$ be the Cantor function. Then $$\phi:J\to{\mathbb R}$$ is continuous and $$\phi'=0$$ on $$U_1\cup U_2\cup U_3\cup\cdots$$ and $$\phi(0)=0$$ and $$\phi(1)=1$$.

For each integer $$k\ge1$$, let $$f_k:{\mathbb R}\to{\mathbb R}$$ be a $$C^\infty$$ function s.t. $$f_k(0)=0$$ and s.t. $$f_k(1)=1$$ and s.t. $$f_k=\phi$$ on $$U_k$$ and s.t. $$\{x\in{\mathbb R}\,|\,f'_k(x)\ne0\}\subseteq U_{k+1}$$ and s.t. $$f'_k\ge0$$ on $${\mathbb R}$$.

Then, because $$U_2,U_3,U_4,\ldots$$ are pairwise-disjoint, it follows, $$\forall x\in{\mathbb R}$$, that there is at most one integer $$k\ge1$$ s.t. $$x\in U_{k+1}$$. Then, $$\forall x\in{\mathbb R}$$, there is at most one integer $$k\ge1$$ s.t. $$f'_k(x)\ne0$$. Then the pointwise limit $$g$$ of $$f'_1,f'_2,f'_3,\ldots$$ is the constant function $$0$$ on $${\mathbb R}$$.

The pointwise (or, even, uniform) limit $$f$$ of $$f_1,f_2,f_3,\ldots$$ is $$0$$ on $$(-\infty,0)$$ and is $$1$$ on $$(1,\infty)$$ and is $$\phi$$ on $$J$$. Then $$f$$ is continuous everywhere on $${\mathbb R}$$. Also, $$f$$ does not have finite derivative at any point of the Cantor set $$C$$.

In particular, we see that $$f'=g$$ fails to be true at any point of $$C$$.

Third, we might only assume that $$f_1,f_2,f_3,\ldots$$ converges pointwise to a differentiable limit $$f$$, keeping the hypothesis that $$f'_1,f'_2,f'_3,\ldots$$ converges pointwise to a continuous limit $$g$$, and then hope to prove continuous differentiability everywhere on $${\mathbb R}$$ of $$f$$. If we could prove such a result, then the theorem above would give: $$f'=g$$.

A counterexample to this is similar to the one in the second comment, but relies on a lot of machinery for finding complicated derivatives. A standard reference for this machinery is Andrew Bruckner's book, "Differentiation of Real Functions". (The Pompeiu derivative is already somewhat difficult, and I sometimes refer to these more complicated methods as "Pompeiu on steroids"!)

The $$f$$ in the counterexample described below is differentiable and satisfies both $$f'\ge0$$ on $${\mathbb R}$$ and $$f'$$ is bounded on $${\mathbb R}$$.

Let $$A:=(-\infty,0]$$. Let $$J:=[0,1]$$. Let $$B:=[1,\infty)$$.

By forming a sufficiently fat Cantor set, and letting $$\phi$$ be the Cantor function of that set, we can guarantee that $$\phi:J\to{\mathbb R}$$ is Lipschitz. ((Proof: Choose a sequence $$\eta_1,\eta_2,\eta_3,\ldots$$ of positive real numbers such that $$\eta_1+\eta_2+\eta_3+\cdots<\infty$$. Start with the identity function on $$J$$. It is Lipschitz-$$1$$. Identify a very small middle interval inside of $$J$$. This splits $$J$$ into three intervals; there are now two larger intervals separated by one very small middle interval. The new Cantor approximation function is constant on the very small interval and slightly steeper than Lipschitz-$$1$$ on the other two. By making the middle interval small enough, the new function is Lipschitz-$$(1+\eta_1)$$. Now identify very, very small middle intervals inside each of the two larger intervals. Take care that the new Cantor approximation function is Lipschitz-$$(1+\eta_1+\eta_2)$$. Continue.))

We have: $$\phi(0)=0$$ and $$\phi(1)=1$$. Then $$\phi$$ is nonconstant. Also, $$\phi$$ is semi-increasing.

Choose pairwise-disjoint closed intervals $$I_1,I_2,I_3,\ldots\subseteq(0,1)$$ all of positive length s.t., $$\forall$$integers $$k\ge1$$, $$\phi$$ is constant on $$I_k$$ and s.t. $$I_1\cup I_2\cup I_3\cup\cdots$$ is dense in $$J$$.

Extend $$\phi$$ to a function $$\psi:{\mathbb R}\to{\mathbb R}$$ satsifying $$\psi=0$$ on $$A$$, $$\psi=\phi$$ on $$J$$, $$\psi=1$$ on $$B$$. Then $$\psi$$ is Lipschitz. Also, $$\forall$$integer $$k\ge1$$, we have: $$\psi$$ is constant on $$I_k$$. Also, $$\psi$$ is constant on $$A$$. Also, $$\psi$$ is constant on $$B$$. Also, $$\psi$$ is nonconstant. Also, $$\psi$$ is semi-increasing. Also, $$A\cup B\cup I_1\cup I_2\cup I_3\cup\cdots$$ is dense in $${\mathbb R}$$.

Let $$Z$$ be the set of points where $$\psi$$ is not differentiable. Since $$\psi$$ is Lipschitz, we conclude that $$Z$$ is a null set, i.e., a set of Lebesgue measure $$0$$. Also, $$\psi'$$ is defined and bounded on $${\mathbb R}\backslash Z$$.

Since $$\phi$$ is Lipschitz, it carries null sets to null sets. Then the image $$Y:=\psi(Z)$$ of $$Z$$ under $$\psi$$ is null. Also, $$\forall x\in Z$$, we have $$\psi(x)\in Y$$.

Combining Theorem 5.5(a) and Theorem 6.5 in Chapter 2 of Andrew Bruckner's book, "Differentiation of Real Functions", we see that, for any null subset $$X$$ of $${\mathbb R}$$, there exists a strictly-increasing, differentiable function $$\rho:{\mathbb R}\to{\mathbb R}$$ s.t. $$\rho'=0$$ on $$X$$ and such that $$\rho'$$ is bounded. ((Proof: Let $$X'\subseteq{\mathbb R}$$ be a null $$G_\delta$$ such that $$X\subseteq X'$$. Then apply Theorem 6.5 with $$E:={\mathbb R}\backslash X'$$. Repace "$$f$$" in Theorem 6.5 with "$$\lambda$$" to avoid confusion with our "$$f$$" defined below. Then $$\lambda=0$$ on $$X'$$ and $$\lambda>0$$ on $${\mathbb R}\backslash X'$$. By Theorem 5.5(a), choose a differentiable function $$\rho:{\mathbb R}\to{\mathbb R}$$ s.t. $$\rho'$$ is bounded and s.t. $$\rho'=\lambda$$. Then $$\rho'=0$$ on $$X'$$ and $$\rho'>0$$ on $${\mathbb R}\backslash X'$$. Then $$\rho$$ is semi-increasing. A function that is semi-increasing but not strictly-increasing must be constant on a nonempty open interval. Since $$X'$$ is null and since $$\rho'>0$$ on $${\mathbb R}\backslash X'$$, we see that $$\rho$$ is not constant on any nonempty open interval. Then $$\rho$$ is strictly-increasing. Since $$X\subseteq X'$$ and since $$\rho'=0$$ on $$X'$$, we see that $$\rho'=0$$ on $$X$$.))

Applying this to $$Y$$, choose a strictly-increasing, differentiable function $$\rho:{\mathbb R}\to{\mathbb R}$$ s.t. $$\rho'=0$$ on $$Y$$ and s.t. $$\rho'$$ is bounded.

Since $$\rho$$ is strictly-increasing and since $$\psi$$ is nonconstant, we get: $$\rho\circ\psi$$ is nonconstant.

Since $$\rho$$ is strictly-increasing and since $$\psi$$ is semi-increasing, we get: $$\rho\circ\psi$$ is semi-increasing.

Since $$\psi$$ is constant on each of: $$A,B,I_1,I_2,I_3,\ldots$$, we get: $$\rho\circ\psi$$ is constant on each of those same sets.

For all $$x\in{\mathbb R}\backslash Z$$, since $$\psi$$ is differentiable at $$x$$ and since $$\rho$$ is differentiable at $$\psi(x)$$, we get: $$\rho\circ\psi$$ is differentiable at $$x$$ and $$(\rho\circ\psi)'(x)=(\rho'(\psi(x))\cdot(\psi'(x))$$. So, since $$\psi'$$ is defined and bounded on $${\mathbb R}\backslash Z$$ and since $$\rho'$$ is defined and bounded on $${\mathbb R}$$, it follows that $$(\rho\circ\psi)'$$ is defined and bounded on $${\mathbb R}\backslash Z$$. That is, $$\rho\circ\psi$$ is differentiable on $${\mathbb R}\backslash Z$$ and that $$(\rho\circ\psi)'$$ is bounded on $${\mathbb R}\backslash Z$$.

For all $$x\in Z$$, we have $$\psi(x)\in Y$$, so $$\rho'(\psi(x))=0$$. For all $$x\in Z$$, since $$\psi$$ is Lipschitz and since $$\rho'(\psi(x))=0$$, it is a basic real-analysis exercise to show: $$\rho\circ\psi$$ is differentiable at $$x$$ and $$(\rho\circ\psi)'(x)=0$$. Then $$\rho\circ\psi$$ is differentiable on $$Z$$ and $$(\rho\circ\psi)'$$ is bounded on $$Z$$.

By the preceding two paragraphs, we get: $$\rho\circ\psi$$ is differentiable and $$(\rho\circ\psi)'$$ is bounded.

Let $$f:=\rho\circ\psi$$. Then $$f$$ is differentiable. Also, $$f$$ is semi-increasing. Also, $$f'$$ is bounded. Also, $$\forall$$integer $$k\ge1$$, $$f$$ is constant on $$I_k$$. Also, $$f$$ is constant on $$A$$. Also, $$f$$ is constant on $$B$$. Also, $$f$$ is nonconstant.

Recall that $$A\cup B\cup I_1\cup I_2\cup I_3\cup\cdots$$ is dense in $${\mathbb R}$$.

For each integer $$k\ge1$$, choose a semi-increasing $$C^\infty$$ function $$f_k:{\mathbb R}\to{\mathbb R}$$ s.t. $$f_k=f$$ on $$A\cup B\cup I_1\cup\cdots\cup I_j$$ and s.t. $$\{x\in{\mathbb R}\,|\,f'_k(x)\ne0\}\subseteq I_{k+1}\cup I_{k+2}\cup I_{k+3}\cup\cdots$$.

Then the pointwise limit $$g$$ of $$f'_1,f'_2,f'_3,\ldots$$ is the constant function $$0$$ on $${\mathbb R}$$.

Also, the pointwise (or, even, uniform) limit of $$f_1,f_2,f_3,\ldots$$ is $$f$$.

Since $$f$$ is nonconstant and $$g$$ is identically zero, we see that $$f'=g$$ fails to be true.

• Thank you so much for the many counterexamples! This answers quite a few questions I had about these as well Jul 21, 2022 at 4:21