Let $\#$ denote cardinality, and fix $p\in[1,\infty]$.
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of functions from $[0,1]$ to $\mathbb{R}$ with the properties

  1. $f_n\in C^{n-1}([0,1])$
  2. $f^{(n)}_n$ is continuous on $[0,1]\setminus A_n$ where $\#A_n<\infty$ but $ \underset{n\to\infty}{\lim}\#A_n=\infty$
  3. $\underset{n\to\infty}{\lim}||f_n^{(m)}||_p$ exists and is finite $\forall m\in\mathbb{N}_0$
  4. $f_n$ converge pointwise to a function $f$

For what values of $p$ do these imply $f\in C^\infty([0,1])$?
For those $p$ where a counterexample exists, how to construct such an example and/or what would be a convenient additional condition to avoid its existence?

Attempt with $p=\infty$:
Assume that $f^{(0)}$ is not continuous. Then there exists $x\in(0,1)$ and $\varepsilon>0$ such that however small $\delta>0$ is, there exists $x_0$ satisfying $0<|x-x_0|<\delta$ such that $|f(x_0)-f(x)|>\varepsilon$. Moreover $0<|(x_0(\delta)-(x+\delta))/2|<\delta$, and if $p=\infty$ then 3) implies $\infty>\underset{n\to\infty}{\lim}|f_n^{(1)}(x)|$. A contradiction follows:

$$\infty>\underset{n\to\infty}{\lim}|3f_n^{(1)}(x)| =\underset{n\to\infty}{\lim}\left(\underset{\delta\to 0^+}{\lim}\left|\frac{f_n(x+\delta)-f_n(x)}{\delta}\right|+2\underset{\delta\to 0^+}{\lim}\left|\frac{f_n(x_0(\delta))-f_n(x+\delta)}{x_0(\delta)-(x+\delta)}\right|\right) \hspace{6.4cm}\text{ }\\\hspace{4cm} >\underset{n\to\infty}{\lim}\left(\underset{\delta\to 0^+}{\lim}\left|\frac{f_n(x+\delta)-f_n(x)}{\delta}\right|+\underset{\delta\to 0^+}{\limsup}\left|\frac{f_n(x_0(\delta))-f_n(x+\delta)}{\delta}\right|\right) \\\hspace{2.15cm} >\underset{n\to\infty}{\lim}\underset{\delta\to 0^+}{\limsup}\left|\frac{f_n(x+\delta)-f_n(x)}{\delta}+\frac{f_n(x_0(\delta))-f_n(x+\delta)}{\delta}\right| \\\hspace{2cm} =\underset{n\to\infty}{\lim}\underset{\delta\to 0^+}{\limsup}\left|\frac{f_n(x_0(\delta))-f_n(x)}{\delta}\right| \hspace{4.6cm}\text{ }\\ \overset{\text{???}}{=}\underset{\delta\to 0^+}{\limsup}\underset{n\to\infty}{\lim}\left|\frac{f_n(x_0(\delta))-f_n(x)}{\delta}\right| \hspace{2.6cm}\text{ }\\ =\underset{\delta\to 0^+}{\limsup}\frac{|f(x_0(\delta))-f(x)|}{\delta} >\underset{\delta\to 0^+}{\limsup}\frac{\varepsilon}{\delta}=\infty \hspace{0.5cm}\text{ }$$

so $f^{(0)}$ is continuous and replacing $f$ and $f_n$ by their derivative smoothness follows by induction. Unless the part with those "???" is wrong?!

  • $\begingroup$ In $3$ is it for all $p\geq 1$ or for some $p\in [1,\infty)$? $\endgroup$
    – clark
    Jan 1, 2018 at 20:24
  • $\begingroup$ @clark Some $p\in[1,\infty]$ $\endgroup$
    – MeMyselfI
    Jan 1, 2018 at 20:26
  • $\begingroup$ In 2. you require that #$A_n\to \infty?$ $\endgroup$
    – zhw.
    Jan 1, 2018 at 21:03
  • $\begingroup$ @zhw. Yes I do. $\endgroup$
    – MeMyselfI
    Jan 1, 2018 at 21:49

1 Answer 1


Actually a stronger result holds:

Thm: Suppose $f_n\in C^{n-1}([0,1]), n =1,2,\dots,$ and $f_n \to f$ pointwise on $[0,1].$ Suppose further that for each $m\in \{0,1,\dots \},$

$$\sup_{n>m+1} \|D^m f_n\|_1 < \infty.$$

Then $f\in C^{\infty}([0,1]).$

Lemma: Suppose $g_n\in C^{2}([0,1]), n =1,2,\dots$ and $\|g_n\|_1,\|g_n'\|_1,\|g_n''\|_1$ are uniformly bounded. Then there exists a subsequence $g_{n_k}$ that converges uniformly on $[0,1].$

To see how the lemma implies the theorem, we start by showing $f\in C^1.$ Consider the sequence $g_n = f_{n+3}', n=1,2,\dots$ From the hypotheses in the theorem, $g_n$ satisfies the hypotheses in the lemma. Thus there exists a subsequence $g_{n_k}$ that converges uniformly to some continuous $g.$

We thus have the following: $f_{n_k+3} \to f$ pointwise, and $f_{n_k+3}'\to g$ uniformly. By the standard result on uniform convergence and differentiation, $f$ is differentiable on $[0,1]$ and $f'= g.$ Since $g$ is continuous, $f\in C^1.$

We now look at $f_{n_k+3}''.$ The tail end of this sequence satisfies the hypotheses of the lemma. Thus some subsequence $f_{n_{k_j}+3}''$ converges uniformly to some continuous $h.$ In exactly the same way as above, we have $f'' = g' =h$ and $f\in C^2.$

Clearly this process can be continued, which proves $f\in C^\infty.$

Proof of the lemma: Suppose $\|g_n\|_1,\|g_n'\|_1,\|g_n''\|_1$ are all bounded above by $C.$ By Fatou's lemma, $\int_0^1 \liminf |g_n'| \le C.$ Thus $\liminf |g_n'| < \infty$ a.e. Thus there exists $b\in [0,1]$ such that for some subsequence $n_k,$

$$\sup_k |g_{n_k}'(b)|=M_b < \infty$$

for all $k.$ Going to a further subsequence $g_{n_{k_j}},$ we will also have $a\in [0,1]$ such that $\sup_k |g_{n_{k_j}}(a)|=M_a < \infty.$

From the FTC it follows that

$$|g_{n_{k_j}}'(x)| \le |g_{n_{k_j}}'(x)-g_{n_{k_j}}'(b)| + |g_{n_{k_j}}'(b)| \le \int_0^1 |g_{n_{k_j}}''| + M_b \le C + M_b.$$

Thus $g_{n_{k_j}}'$ is uniformly bounded on $[0,1].$ By the MVT, $g_{n_{k_j}}$ is uniformly Lipschitz on $[0,1].$ Thus the sequence $g_{n_{k_j}}$ is equicontinuous on $[0,1].$ Since $g_{n_{k_j}}(a)$ is bounded, Arzela-Ascoli implies there is a subsequence of $g_{n_{k_j}}$ that converges uniformly, which gives the lemma.


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