# Is the $W^{1, \infty}$ limit of differentiable functions also differentiable?

Let $$f_n$$ be a sequence of differentiable functions on $$[0, 1]$$ with

1. $$f_n \to f$$ uniformly for some (necessarily) continuous $$f$$.
2. $$f'_n - g \to 0$$ in $$L^{\infty}$$ for some measurable $$g$$.

Is it true that $$f$$ is differentiable everywhere, with $$f' = g$$ almost everywhere?

Some comments: An almost everywhere version of the same question is answered negatively here. I expected to have an easy affirmative answer if $$f$$ are assumed instead everywhere differentiable, but to my surprise this seems to be much more subtle. Note that we do not assume that $$f’_n$$ are in $$L^1$$, nor that $$f_n$$ are absolutely continuous, so that the fundamental theorem of calculus does not apply.

Update:

The answer by CityHunter below shows that we have that $$f$$ is differentiable a.e. with $$f’ = g$$ a.e., under the weaker condition that $$f’_n - g \to 0$$ in $$L^1$$. Thus the remaining question, and the one I expect to be hardest is the question of whether $$f$$ is differentiable everywhere. To isolate the essential difficulty of the problem of everywhere differentiability, we could maybe consider first the following version with stronger convergence assumptions.

Assume $$f_n \to f$$ in $$W^{1, \infty}$$ with $$f_n$$ everywhere differentiable. Then is $$f$$ everywhere differentiable?

Even this weaker version seems to be nontrivial.

Update 2: It seems the weaker version is true, so it is left to either remove the regularity requirements as in the original post, or to try to derive a multidimensional version of the latter claim. I have added an outline of a proof strategy for the multidimensional case in the answers below.

• If anyone is willing to work with me more closely on this, perhaps we could start a MathOverflow chat room for the problem. Commented May 26 at 1:12
• count me in!  Commented Jun 6 at 9:45
• @PietroMajer Sorry that the email is taking awhile, I want to have some meaningful progress to share before I send anything! I am getting closer to filling in the hard details for the multidimensional case. Although I am working on the limit formulation, but I think they should be equivalent. I don’t see how to show that immediately though… Commented Jun 9 at 2:19
• Take your time! ;) Commented Jun 9 at 6:18
• @PietroMajer Done! Since I seem to have managed a full proof of the multidimensional case, I wrote it in full below instead of emailing. It seems to be almost too easy though, so I am skeptical that I got it right... But if this works then we can explore the $\mathcal H^k$ version. :D Commented Jun 10 at 5:23

This is true, $$f$$ is actually everywhere differentiable. It is the "Limit under the Sign of Derivative" Theorem; it also holds for sequences of maps between Banach spaces (and you may even allow countably many points of non-differentiability). Check e.g. Dieudonné's Foundations of Modern Analysis , Theorem $$(8.6.3)$$ .

Edit. As immediately spotted by Josif Pinelis, to apply the LSD Theorem as stated by Dieudonné, we need to replace the uniform convergence on a full measure set (2) with the global (also "local" would go) uniform convergence, modifying $$g$$ on a null set if needed. This is possible because of:

Prop 1. For an everywhere differentiable function $$u:[a,b]\to \mathbb R$$ one has $$\text{sup ess } |u'|=\sup|u'|$$ So this is another "fake continuity" property of functions that are derivatives of everywhere differentiable functions, in the spirit of Darboux theorem.

In particular the sequence $$(f_n)$$ has $$\text{sup ess } |f_p'-f_q'|=\sup| f_p'-f_q'|$$ for all $$p,q$$ and since by the assumption (2) $$(f_n')_n$$ is Cauchy in $$L^\infty$$, it is also Cauchy w.r.to uniform convergence. So by the LSD Thm it converges uniformly to a function $$\tilde g$$ which is the everywhere derivative of $$f$$, and coincides a.e. with $$g$$.

Since $$\text{ sup ess } |u'(x)|\le\sup|u'(x)|=\sup_{a\le x the above Prop 1 follows from a Mean Value Theorem for everywhere differentiable functions (applied on each interval $$[x,y]$$) :

Prop 2. Assume $$u:[a,b]\to \mathbb R$$ is everywhere differentiable. Then $$\frac{u(b)-u(a)}{b-a}\le\text{ sup ess } u'(x) .$$

(a fact that is notoriously false if one only assumes $$u$$ continuous and differentiable a.e.). To prove Prop 2 we consider $$v(x):=u(x)-\frac{u(b)-u(a)}{b-a}(x-a)$$: it is sufficient to prove $$\text{ sup ess }v'(x) \ge0$$, which follows from

Prop 3. Assume $$v:[a,b]\to \mathbb R$$ is everywhere differentiable and $$v'(x)\le0$$ a.e. Then $$v$$ is decreasing.

I wish to prove a slightly stronger version of Prop 3, namely [...]

[edit 05.30.24]

...BUT I'll better do in another answer [edit: Now available below!] as this is becoming too long. In the meanwhile here is the

Proof of Prop 3. Assume by contradiction $$v$$ is non-decreasing, say w.l.o.g. $$v(b)>v(a)$$. The function $$v(x)-\frac12\frac{v(b)-v(a)}{b-a}x$$ has the same property, and a derivative negative a.e. So we shall make the stronger assumption that $$v$$ is everywhere differentiable with $$v'(x)<0 \;\text{ a.e. and } \; v(b)>v(a).$$

The intervals $$[c,d]\subset(a,b)$$ with $$v(d) cover the set $$\{v'<0\}$$ in the Vitali sense, that is, every point of $$\{v'<0\}$$ belongs to arbitrarily small such intervals. By the Vitali covering lemma, there is a finite disjoint family of these intervals whose sum of lengths is greater than $$\frac12(b-a)$$. In other words, labelling these intervals $$[c_{2k-1},c_{2k}]$$ for $$k=1\dots n$$, there exists a finite sequence $$c_0:=a such that $$v(c_{2k}) and $$\sum_{k=0}^n(c_{2k+1}-c_{2k})=(b-a)- \sum_{k=1}^n(c_{2k}-c_{2k-1}) \le \frac12(b-a).$$ Therefore $$v(b)-v(a)= \sum_{j=0}^{2n} v(c_{j+1})-v(c_j)\le \sum_{k=0}^nv(c_{2k+1})-v(c_{2k})=$$ $$=\sum_{k=0}^n\frac{v(c_{2k+1})-v(c_{2k})}{c_{2k+1}-c_{2k}}(c_{2k+1}-c_{2k})\le \sum_{k=0}^n(c_{2k+1}-c_{2k})\max_{0\le k\le n} \frac{v(c_{2k+1})-v(c_{2k})}{c_{2k+1}-c_{2k}} =$$ $$\le\frac12(b-a)\max_{0\le k\le n} \frac{v(c_{2k+1})-v(c_{2k})}{c_{2k+1}-c_{2k}}$$

That is, for the maximising index $$k^*$$, the interval $$[c,d]:=[c_{2k^*+1},c_{2k^*}]$$ has $$\frac{v(d)-v(c)}{d-c} \ge2 \frac{v(b)-v(a)}{b-a}.$$ If we iterate this procedure we get a nested sequence $$[a_n,b_n]\subset [a,b]$$ with $$\frac{v(b_n)-v(a_n)}{b_n-a_n}\to\infty.$$ If $$x_*\in\bigcap_{n\ge0}[a_n,b_n]$$, we have $$\limsup_{x\to x^*}\frac{v(x^*)-v(x)}{x^*-x}=+\infty$$, a contradiction.

Rmk. Note that in the above proof, the everywhere differentiability of $$v$$ was used just to reach the contradiction. To put it in a positive form, by the same argument we have: Assume $$v:[a,b]\to\mathbb R$$ is continuous, with lower Dini derivative $$D_*v(x)\le0$$ a.e., and $$v(a)>v(b)$$. Then there is a point $$x^*\in[a,b]$$ with infinite upper Dini derivative: $$D^*v(x^*)=+\infty$$. It is clear that we can apply this result locally, so that actually there are infinitely many points where $$D^*v=+\infty$$. Moreover in the construction of the intervals $$[\alpha_n,\beta_n]$$ we can skip every point in a given countable subset as limit, so that actually the set $$\{D^*v=+\infty\}$$ is uncountable. Or better: if at the $$n+1$$-th step we pick two disjoint closed intervals within every interval of the $$n$$-th step, we get a whole Cantor set $$K\subset\{D^*v=+\infty\}$$, thus of cardinality $$\bf\mathfrak c.$$ This is what happens e.g. for the Cantor-Vitali function, for which $$C=\{D^*v=+\infty\}$$ is exactly the triadic Cantor set. This example suggest that in general one should have $$v(\{D^*v=+\infty\})\supset [v(a),v(b)]$$, at least up to negligible sets. I'd try to post a proof of this soon.  Done! See below.

• It's a fantastic book from the roaring 60's Commented May 23 at 17:13
• In Theorem (8.6.3), $f_n'$ is required to converge locally uniformly, but that does not seem to easily follow from the conditions in the OP. In particular, the second condition in the OP only provides the uniform convergence of $f_n'$ on a set of full measure. Commented May 23 at 21:15
• @NateRiver: Pietro's "claim" that the esssup |f'| = sup |f'|, if true, would imply that the scenario you envision is impossible, as it would imply that $f_n'$ is Cauchy under the uniform norm. Conversely, if you have a specific example of a sequence $f_n$ satisfying your hypotheses and a point $x$ at which $(f_n'(x))$ is not Cauchy, then it would disprove Pietro's conjecture. Commented May 24 at 1:51
• As soon as the sun rises Commented May 24 at 3:20
• It is a very interesting question and it made me think about some old facts which are still a source of curiosity, so it's me who has to thank you :) Commented May 31 at 5:37

It suffices to assume $$f’_n-g\overset{L^1}{\to}0$$.

Consider large $$N$$ such that $$||f’_n-g||_{L^1}<\infty~(\forall n\ge N)$$, put $$h_n=f_n-f_N$$ and $$h=f-f_N$$. We have that $$h_n\overset{unif}{\to}h$$ and $$h’_n \overset{L^1}{\to} g-f’_N$$.

Since $$h’_n$$ exists everywhere and is $$L^1~(\forall n\ge N)$$ the Fundamental Theorem of Calculus holds:

$$\frac{h(x)-h(y)}{x-y}=\lim_{n\to\infty}\frac{h_n(x)-h_n(y)}{x-y}\\~~~~~~~~~~~~~~=\lim_{n\to\infty}\frac{1}{x-y}\int^x_y h’_ndt\\~~~~~~~~~~~~~~=\frac{1}{x-y}\int^x_y g-f’_Ndt$$

Since $$g-f’_N\in L^1$$ we have $$h’=g-f’_N$$ at every Lebesgue point of $$g-f’_N$$ hence almost everywhere. Finally since $$f’_N$$ exists everywhere hence whenever $$h’=(f-f_N)’$$ exists $$f’$$ must also exist, and so

$$f’=(f’-f’_N)+f’_N=h’+f’_N=g$$ almost everywhere, as desired.

• Nice! I believe this derivation almost resolves the problem - it shows that $f$ must be differentiable a.e. with $f’ = g$ a.e. It is only left to deal with the tricky problem of whether $f$ is in fact differentiable everywhere. Commented May 24 at 9:00
• To isolate the essential difficulty of the everywhere differentiability problem, we could consider instead the case with $f’_n, g \in L^\infty$. The question of whether $f$ is differentiable everywhere is still not trivial in this case I believe. Commented May 24 at 9:23
• (Of course since $C^\infty(I)$ is dense in $W^{1,1}(I)$, in the case of $L^1$ convergence, $f$ may fail to be everywhere differentiable) Commented May 30 at 6:24

Taken by a sort of generalisation frenzy I produced the following; I tried to make it as readable as I could. Recall that the upper and lower Dini derivatives are respectively : $$D^*f(x):=\limsup_{y\to x}\frac{f(y)-f(x)}{y-x}$$ $$D_*f(x):=\liminf_{y\to x}\frac{f(y)-f(x)}{y-x}.$$

Theorem. Let $$f:[a,b]\to \mathbb R$$ be a continuous function with $$D_*f\le0$$ a.e. Then $$f(b)-f(a)\le \big|f(\{D^*f=+\infty\})\big|$$

In particular, if $$D_*f\le0$$ a.e. and $$D^*f<\infty$$ everywhere, applying this to every sub-interval one has that $$f$$ is decreasing.

The archeotypical example for this situation is the Cantor function, that has $$f'(x)=0$$ a.e. and $$f(\{D^*f=+\infty\})=[f(0), f(1)]=[0,1]$$, because $$\{D^*f =+\infty\}$$ is exactly the triadic Cantor set $$C$$. In general, the situation is slightly more complicated, but keeping this example in mind, we can construct, for given $$\epsilon>0$$, a Cantor set $$K= \bigcap_{n\ge0} K_n\subset \{D^*f=+\infty\}$$ such that $$\big|[f(a),f(b)]\setminus f(K)\big|\le \epsilon$$. Here the sets $$K_n$$ are finite unions of closed intervals, or pluriintervals, inductively defined.

It is convenient to denote, for a pluriinterval $$K\subset[a,b]$$ with connected components $$\big\{[\alpha_j,\beta_j]\big\}_{1\le j\le p}$$, that is a disjoint union $$\displaystyle K:=\bigsqcup_{1\le j\le p} [\alpha_j,\beta_j]$$, $$K^f: =\bigcup_{1\le j\le p} [f(\alpha_j),f(\beta_j)].$$ (Here $$[x,y]=\emptyset$$ if $$x>y$$). Note that $$K^f\subset f(K)$$ by continuity of $$f$$.

Rmk. In what follows it may be useful to recall the elementary $$A\setminus C\subset B\Leftrightarrow A\subset B\cup C$$.

For the proof we need a Lemma. In the same hypotheses of the Theorem we have

Lemma. For every $$\eta>0$$, and for every pluriinterval $$K\subset[a,b]$$ there exists a pluriinterval $$L \subset K$$ such that $$\big|K^f \setminus {L}^f\big|\le\eta$$, and $$\frac{f(\beta)-f(\alpha )}{\beta -\alpha }\ge\frac1\eta$$ for every component $$[\alpha,\beta]$$ of $$L$$.

Proof of the Theorem. We define inductively a nested sequence of pluriintervals $$K_n$$ where $$K_0=[a,b]$$ and for all $$n\ge1$$, $$K_n$$ is the pluriinterval $$L$$ given by the Lemma, corresponding to $$K:=K_{n-1}$$, and to the number $$\eta:=\epsilon 2^{-n}$$. We then define $$K:=\bigcap_{n\ge0}K_n$$ and $$S:= \bigcup_{n\ge1} (K_{n-1}^f\setminus K_n^f)$$, which is a set of measure $$|S|\le\epsilon$$ because for all $$n$$ we have $$|K_{n-1}^f\setminus K_n^f|\le \epsilon 2^{-n}$$ .

For every $$n$$ we have $$[f(a),f(b)]\setminus K_n^f= K_0^f\setminus K_n^f \subset S$$, so $$[f(a),f(b)]\setminus S\subset K_n^f\subset f(K_n)$$ and then $$[f(a),f(b)]\setminus S\subset \bigcap_{n\ge0} f(K_n)$$ ; finally since $$K_n$$ is a decreasing sequence of compact sets and $$f$$ is continuous, $$\bigcap_{n\ge0} f(K_n)= f\big( \bigcap_{n\ge0} K_n\big)=f(K)$$. We thus have $$[f(a),f(b)]\subset f(K)\cup S$$ whence $$f(b)-f(a)\le |f(K)|+\epsilon$$

Note that each component interval of $$K_n$$ has length at most $$2^{-n+1}\epsilon\|f\|_\infty$$. So if $$x\in K$$, there is a sequence $$x_n\in K_n$$ of endpoints of components, such that $$x_n\to x$$ with $$\frac{f(x)-f(x_n)}{x-x_n}\ge \frac{2^n}{\epsilon}$$, that is $$K\subset \{D^*f=+\infty\}.$$
Since $$\epsilon>0$$ is arbitrary, the thesis follows.

Rmk. More precisely we have shown the inequality w.r.to the inner measure of the set $$f( \{ D^*f=+\infty\})$$, without addressing the issue of measurability of it. Incidentally, this is sufficient for the OP's needs. In fact $$\{ D^*f=+\infty\}$$ is a $$G_\delta$$ set, and a continuous image of a $$G_\delta$$ set is even a Borel set, by a non-trivial result in Analytic Set Theory (I think for this case there is a direct simpler proof).

Proof of the Lemma. We first do the case $$p=1$$ and $$K:=[a,b]$$. The family of intervals $$[\alpha,\beta]\subset(a,b)$$ such that $$\frac{f(\beta )-f(\alpha )}{\beta -\alpha }<\frac{\eta}{2(b-a)}$$ cover the set $$\{D_*f\le0\}$$ in the Vitali sense, that is, every point of $$\{D_*f\le0\}$$ belongs to arbitrarily small such intervals. By the Vitali covering theorem, there is a finite disjoint family of these intervals whose sum of lengths is larger than $$b-a -\frac{\eta^2}2$$. Thus, labelling these intervals $$[c_{2k-1},c_{2k}]$$ for $$k=1\dots n$$, with a finite sequence $$c_0:=a we have , for $$k=1,\dots,n$$

$$f(c_{2k})-f(c_{2k-1})\le \frac{\eta}{2(b-a)}(c_{2k}-c_{2k-1})$$ and $$\sum_{k=0}^n(c_{2k+1}-c_{2k})=(b-a)- \sum_{k=1}^n(c_{2k}-c_{2k-1})\le \frac{\eta^2}2.$$ Let $$J:=\Big\{k\in \mathbb N:0\le k\le n,\, \frac{f(c_{2k+1})-f(c_{2k})}{c_{2k+1}-c_{2k}} \ge\frac1\eta\Big\}$$ and $$L:=\bigcup_{k\in J} [c_{2k},c_{2k+1}]$$ Then we have

$$[a,b]^f \setminus {L}^f\subset \bigcup_{ k\notin J}[c_{2k} , c_{2k+1}]^f\cup \bigcup_{1\le k\le n} [c_{2k-1} , c_{2k}]^f$$ so $$\bigg|[a,b]^f \setminus {L}^f\bigg|\le \sum_{ k\not\in J}\big( f(c_{2k+1})-f(c_{2k})\big)_+ +\sum_{k=1}^{n}\big( f(c_{2k})-f(c_{2k-1})\big)_+ \le$$ $$\le\sum_{ k=0}^n \frac1\eta \big( c_{2k+1}-c_{2k})+ \sum_{k=1}^{n}\frac{\eta}{2(b-a)}(c_{2k}-c_{2k-1})\le$$

$$\le \frac1\eta \frac{\eta^2}2+\frac{\eta}{2(b-a)}(b-a)=\eta,$$ so $$L$$ enjoyes the property we wished, for the case $$K:=[a,b]$$.

For the case of a general pluri-interval $$\displaystyle K:=\bigsqcup_{1\le j\le p} [\alpha_j,\beta_j]$$, we apply the preceding case $$n=1$$ to each component interval $$[\alpha_j,\beta_j]$$ w.r.to the number $$2^{-j}\eta$$ and take $$L$$ to be the union of the corresponding pluriintervals $$L_j\subset [\alpha_j,\beta_j]$$, for $$j=1,\dots p$$. Then $$K^f \setminus {L}^f = \bigcup_{1\le j\le p} \big([\alpha_j,\beta_j]^f \setminus {L}^f\big) \subset \bigcup_{1\le j\le p} \big( [\alpha_j,\beta_j] ^f\setminus {L_j}^f\big)$$ has measure $$\bigg|K^f \setminus {L}^f\bigg|\le\sum_{1\le j\le p} \big| [\alpha_j,\beta_j] ^f\setminus {L_j}^f\big| \le \sum_{j=1}^p 2^{-j}\eta<\eta,$$ completing the proof.

• Upon rereading the Cantor function example, I realise this result is even cooler than I thought. Commented Jun 4 at 11:47
• I wonder if equality is achieved if $Df = 0$ a.e.? Or even iff… if you stipulate that the inequality hold for every interval. Commented Jun 4 at 11:49
• The equality can be achieved even if $D_*f<0$ a.e. For instance if $c(x)$ is the Cantor function, one can take $f(x): =2c(x)-x$, so $f’=-1$ a.e., $f([0,1])=[0,1]=f(\{D_*f=+\infty\})$ Commented Jun 4 at 12:26
• The Cantor function allows nice examples. Since $c(x)\le x^{\frac{\log3}{\log2}}$ one has $c(x)^2\le x^{\frac{\log4}{\log2}}=o(x)$ as $x\to0$. So the (odd) function $f(x)=c(|x|)^2\text{sgn}(x)$ on $[-1,1]$ has $f’(0)=0$ and $f(\{D_*f=\infty\})=[f(-1), f(1) ]\setminus\{0\}$ . And this way one can easily miss even a countable set of points of $[f(-1), f(1) ]$ Commented Jun 4 at 13:06

I believe I have a proof of the claim for functions in $$\mathbb R^n$$. For convenience, we restate the theorem below.

Theorem: Suppose $$f_n - f \to 0$$ uniformly and $$f’_n - g \to 0$$ in $$L^\infty$$. Then if $$f_n$$ are differentiable, so is $$f$$.

Below I outline a sketch of a proof, which I will progressively turn into a proper proof over the course of the next few days. Any preliminary comments are greatly appreciated! (Done, finally!)

We show, as in Pietro Majer's answer that for an everywhere differentiable function $$u$$ on an convex, open subset $$\Omega$$ of $$\mathbb R^n$$ that

$$\text{esssup}_{\Omega} |u'|=\sup_{\Omega} |u'|.$$

The conclusion then follows from the "Limit under the Sign of Derivative" Theorem, which is Theorem (8.6.3) in Dieudonné's Foundations of Modern Analysis.

Since

$$\sup_{\Omega} |u'| \leq \text{Lip}(u),$$

it will suffice to show that $$u$$ is Lipschitz continuous with Lipschitz constant $$\|u'\|_{L^\infty}$$.

To see this, let $$x, y \in \Omega$$ be arbitrary. By Fubini's theorem, for every $$\varepsilon > 0$$, there exist points $$x_0, y_0$$ that are $$\varepsilon$$-close to $$x, y$$ respectively such that, denoting by $$w$$ the unit vector pointing from $$x_0$$ to $$y_0$$, we have that $$F (t) := (u)(x_0 + tw)$$ satisfies $$F'(t) \leq \|u'\|_{L^\infty}$$ almost everywhere (with respect to one dimensional Lebesgue measure).

In particular since $$F$$ is also differentiable everywhere, the one dimensional result in Proposition 1 of Pietro Majer's answer applies, and so we have

$$u(x_0) - u(y_0) \leq \|u\|_{L^\infty} |x_0 - y_0|.$$

Consequently by the triangle inequality,

$$|u (x) - u(y)| \leq |\|u\|_{L^\infty}|x_0 - y_0| +|u(x) - u(x_0)|$$ $$+ |u(y) - u (y_0)|$$ $$\leq |\|u\|_{L^\infty} (|x - y| + 2\varepsilon) +|u(x) - u (x_0)|$$ $$+ |u(y) - u (y_0)|.$$

Now let $$\delta > 0$$ be arbitrary. Assume that $$\varepsilon$$ is chosen smaller than $$\delta |x - y|$$, and small enough such that by continuity,

$$|u(x) - u (x_0)|, |u(y) - u (y_0)| < \delta|x- y|.$$

Then

$$|u (x) - u(y)| \leq (\|u\|_{L^\infty} + 4 \delta)|x - y|,$$

and sending $$\delta$$ to $$0$$, we obtain the desired Lipschitz continuity. $$\square$$.

• In the case $n=1$ it is obvious that for $f\in W^{1,1}(I)$ one has $A:=\sup_I |f’|=B:=\text{supess}_I|f’|$, isn’t it? Because integrating we have that $f$ is $B$-Lipschitz, so all incremental ratios and the derivative everywhere are bounded by $B$, and $A=B$. Commented Jun 4 at 8:48
• @PietroMajer Ah yes, I agree, the true generalisation to multiple dimensions would involve no restrictions on $f$ other than continuous and differentiable everywhere. (Or $\mathcal H^k$ almost everywhere, as in the new post!) Commented Jun 4 at 9:33
• Yes, e.g. proving sup ||Df|| =supess ||Df|| in multiple dimension seems like another not easy question Commented Jun 4 at 9:59
• Challenging, the version with $\mathcal H^k$ ! Commented Jun 4 at 10:03
• @StefanKohl As a suggestion, I wonder if there could be an option for "minor edits" that don't bump the thread to the front page? Commented Jun 10 at 8:19