# When does the $f^{(n)}$ converge to a limit function as $n\to\infty$?

(This was something someone (almost) asked in a comment in a thread about repeated differentiation of polynomials.)

Consider a general smooth (that is $C^\infty$) function $f$. As usual $f^{(n)}$ denotes the $n$th derivative of $f$.

Under what circumstance does $\{ f^{(n)} \}_{n=1}^\infty$ converge to a limit function as $n$ goes to infinity?

For example when $0<k<1$ is fixed and $f(x)=e^{kx}$, then we have the pointwise convergence $f^{(n)} \to 0$ for $n\to\infty$ where $0$ is the zero function. The case $k=1$ is different.

Of course when $f(x)=\cos x$, there is no convergence of the $f^{(n)}$. But $\cos(kx)$ ...

Can a criterion be given?

• What type of convergence? What domain is $f$ given?
– Paul
Commented Aug 30, 2017 at 19:45
• @Paul Pointwise convergence, for example. Or uniform, if possible. The domain could be an open interval in $\mathbb{R}$, for example. Feel free to interpret. If a particular choice of prerequisites leads to a nice answer, use it. Commented Aug 30, 2017 at 19:50
• If you were in a complete space where every Cauchy sequence converges, the Cauchy sequence would be asymptotically a differential equation which would only allow limit functions of zero or of the form $ae^x$.
– Paul
Commented Aug 30, 2017 at 22:50
• @Paul Nice, and related to the partial answer by nicomezi where they show that under the hypothesis that $f$ be analytical, the limit function must have the form $g(x)=le^x$ for a constant $l$ (which was $l=0$ in my examples above). Commented Aug 31, 2017 at 7:12

Let $f$ a real $C^\infty$ function s.t. $f$ can be expressed as its Maclaurin series with radius of convergence $R$ : $$f(x)=\sum_{k=0}^{\infty}\frac {f^{(k)}(0)}{k!}x^k, \forall x \in (-R,R).$$

By properties of power series, in the interior of the domain of convergence, we have : $$f^{(n)}(x)=\sum_{k=n}^{\infty}\frac {f^{(k)}(0)}{(k-n)!}x^{k-n}=\sum_{k=0}^{\infty}\frac {f^{(n+k)}(0)}{k!}x^{k} , \forall x \in (-R,R).$$

We claim the following :

If : $$\lim_{n \to \infty} f^{(n)}(0)=l ,l\in \mathbb{R}\qquad (*)$$ Then $(f^{(n)})$ converges uniformly to $g(x)=l e^x$ on $(-r,r)$ for any $r$ s.t. $0<r<R$.

Proof

$$|f^{(n)}(x)-le^x|=\left| \sum_{k=0}^{\infty}\frac {f^{(n+k)}(0)}{k!}x^{k}-\sum_{k=0}^{\infty}\frac {l}{k!}x^{k}\right|$$

Then :

$$|f^{(n)}(x)-le^x| \leq \sum_{k=0}^{\infty}\left|\frac {f^{(n+k)}(0)-l}{k!}\right|r^{k}.$$

Since we have $(*)$, let $\varepsilon>0$, for $n$ big enough : $$|f^{(n)}(x)-le^x|\leq \varepsilon e^r.$$

Since $x$ is arbitrary, the claim follows.

Remark:

I believe nothing more can be done with less constraints on the sequence $(f^{(n)}(0))$. For example, if you suppose it only bounded, then consider $f(x)=\cos(x)$.

• The only reason why your answer is only partial, as far as I can see, is because you disregard the case where $f$ is $C^\infty\setminus C^\omega$, i.e. smooth but not analytic. On your Remark: Of course, if we want $f^{(n)}$ to be convergent as a sequence of functions, in most senses, then automatically $f^{(n)}(0)$ is convergent as a sequence of numbers, and we can just call the limit $l$. Commented Aug 31, 2017 at 7:19
• Indeed. Now I wonder if there is simple examples of analytic functions where $l \neq 0$. (Apart from the trivial $f(x)=le^x$.) By simple, I mean with a closed form in terms of usual functions. Commented Aug 31, 2017 at 9:19