Let $f :\mathbb R \to\mathbb R$ a function $ n \geq 2$ times differentiable so that: $ \lim_{x \to \infty} f(x) = l \in \mathbb R$ and $ \lim_{x \to \infty} f^{(n)}(x) = 0$. Prove that: $ \lim_{x \to \infty} f^{(k)}(x) = 0 $ for all $ k \in \{1, 2, \dots, n - 1\} $, where $f^{(k)}$ is the $ k $ - th derivative of $f$.
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Tags: real analysis, Taylor expansion, derivative
28.04.2021 21:51
By Taylor's theorem for each $1 \le k \le n-1$ there exists $y_k \in (x,x+k)$ such that $$\sum_{j=1}^{n-1}\frac{f^{(j)}(x)}{j!}k^j=f(x+k)-f(x)-\frac{f^{(n)}(y_k)}{n!}k^n,$$which can be written as $$\begin{pmatrix}1 & 1 & \cdots & 1 \\ 2 & 2^2 & \cdots & 2^{n-1} \\ . & . & \cdots & . \\ . & . & \cdots & . \\ . & . & \cdots & . \\ n-1 & (n-1)^2 & \cdots & (n-1)^{n-1} \end{pmatrix}\begin{pmatrix}f'(x) \\ \frac{f''(x)}{2} \\ . \\ . \\ . \\ \frac{f^{(n-1)}(x)}{(n-1)!}\end{pmatrix}=\begin{pmatrix}g_1(x) \\ g_2(x) \\ . \\ . \\ . \\ g_{n-1}(x)\end{pmatrix},$$where $$g_j(x):=f(x+j)-f(x)-\frac{f^{(n)}(y_j)}{n!}j^n, \ \ \ 1 \le j \le n-1.$$Since the square matrix on the LHS of the above is invertible (Vandermonde), we get that each $f^{(k)}(x), \ \ 1 \le k \le n-1,$ is a linear combination of $g_1(x), \cdots , g_{n-1}(x).$ So, in order to prove that $\lim_{x\to\infty}f^{(k)}(x)=0$ for all $1 \le k \le n-1,$ we only need to show that $\lim_{x\to\infty}g_j(x)=0$ for all $1 \le j \le n-1$ and that's easy to see. For each $j,$ we have $\lim_{x\to\infty}(f(x+j)-f(x))=0$ because we're given that $\lim_{x\to\infty}f(x)$ is a finite number. Also, $\lim_{x\to\infty}f^{(n)}(y_j)=0$ because we're given that $\lim_{x\to\infty}f^{(n)}(x)=0$ and $x < y_j < x+j.$ Thus $\lim_{x\to\infty}g_j(x)=0$ for all $j.$
29.04.2021 03:45
very nice solution
29.04.2021 06:46
Nice....
02.05.2021 12:07
It's natural to use Taylor's expansion. $$f(a+t)=f(a)+f'(a)\frac{t}{1!}+\dots +f^{(n-1)}(a)\frac{t^{n-1}}{(n-1)!}+f^{(n)}(\xi)\frac{t^n}{n!}$$where $\xi\in (a,a+t).$ It means $$\frac{f'(a)}{1!}t+\dots +\frac{f^{(n-1)}(a)}{(n-1)!}t^{n-1} = f(a+t)-f(a)-f^{(n)}(\xi)\frac{t^n}{n!}\qquad (1)$$We take $a$ large enough and vary $t$ in the interval $[0,1].$ So, the RHS of $(1)$ can be made as small as we want as $a\to\infty$ and we want to show all $f^{(k)}(a), k=1,2,\dots,n-1$ are also small. The following lemma shows why it should hold. It has also a standalone importance. For example, the idea behind it was used in Putnam 1999, A5. Lemma. Let $r\in \mathbb{N}.$ There exists a constant $C$ (depending only on $r$) such that for any polynomial $P(t)=a_0t^r+a_1t^{r-1}+\dots a_{r-1}t+a_r$ we have $$|a_0|+|a_1|+\dots+|a_r|\le C\max_{t\in[0,1]} |P(t)|$$Proof. Consider the vector space $L_r$ of all polynomials of degree at most $r.$ It can be normed by any of the following two norms $$\|p\|_1:= |a_0|+|a_1|+\dots+|a_r|\,;\, \|p\|_2:=\max_{t\in[0,1]}|p(t)|$$for any $p\in L_r, p(t)=a_0t^r+a_1t^{r-1}+\dots a_{r-1}t+a_r$. Since $L_r$ is a finite dimensional vector space, any two norms are equivalent, which means there exist constants $C_1,C_2>0$ depending eventually only on $r$ so that $$C_1\|p\|_2\le \|p\|_1\le C_2\|p\|_2,\forall p\in L_r$$which completes the proof. $\blacksquare$ Back to the original problem. Taking into account $(1)$ and the given conditions, for any $\varepsilon>0$ we can take large enough $A>0$ such that for all $a>A$ and $t\in[0,1]$ it holds $$\left|\frac{f'(a)}{1!}t+ \frac{f''(a)}{2!}t^2 +\dots +\frac{f^{(n-1)}(a)}{(n-1)!}t^{n-1}\right|<\varepsilon$$By the above lemma we get $$\frac{\left|f'(a)\right|}{1!} +\frac{\left|f''(a)\right|}{2!}+\dots +\frac{\left|f^{(n-1)}(a)\right|}{(n-1)!}<C\varepsilon$$where $C$ is a constant, depending only on $n$. It implies $$\lim_{a\to\infty}f^{(k)}(a)=0, k=1,2,\dots,n-1.$$ Remark. Some other properties of polynomials, similar to that used in the above lemma.
21.05.2021 10:40
I'm one of the contributors to this problem
22.05.2021 19:22
I'm pretty sure this is an old problem. It was solved in a SEEMOUS training session in my university last year.