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

Question

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?

Answer 1

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 $|\phi'(q)|>\varepsilon$.

Contradiction. QED

Answer 2

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 $f'=g$.

Three comments:

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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 $0$.

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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 $C$.

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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.