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For all $t>0$, define $f_0(t) = \dfrac{1}{e^t-1}$ and $f_{n+1}(t)=\dfrac{3}{4t} + \dfrac{t^3}4 (f_n(t))^4.$

Does this sequence converge uniformly to $\dfrac 1t$ on the positive real numbers?

Remarks: It is not hard to see that the sequence converges monotonously to $\dfrac 1t$. Therefore, Dini's theorem guarantees uniform convergence on compact intervals, and it is not hard to extend this towards infinity, as all functions tend to zero there. However, near zero all the functions have a singularity, so it is unclear to me what happens to the difference at/near $0$.

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  • $\begingroup$ I think the word you want is 'monotonically'; the work itself may be monotonous, but the sequence probably isn't! :-) $\endgroup$ – Steven Stadnicki Feb 10 '14 at 19:23
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Let us write

$$d_n(t) = \frac{1}{t} - f_n(t).$$

Then we find

$$\begin{align} f_{n+1}(t) &= \frac{3}{4t} + \frac{t^3}{4}f_n(t)^4\\ &= \frac{3}{4t} + \frac{t^3}{4}\left(\frac{1}{t} - d_n(t)\right)^4\\ &= \frac{3}{4t} + \frac{t^3}{4}\left(\frac{1}{t^4} - \frac{4}{t^3}d_n(t) + \frac{6}{t^2}d_n(t)^2 - \frac{4}{t}d_n(t)^3 + d_n(t)^4\right)\\ &= \frac{1}{t} - d_n(t) + \frac{3}{2}t\, d_n(t)^2 - t^2d_n(t)^3 + \frac{t^3}{4} d_n(t)^4, \end{align}$$

so

$$d_{n+1}(t) = d_n(t) - \frac{3}{2}t\,d_n(t)^2 + t^2d_n(t)^3 - \frac{t^3}{4} d_n(t)^4.\tag{1}$$

Starting with

$$\begin{align} d_0(t) &= \frac{1}{t} - \frac{1}{e^t-1}\\ &= \frac{1}{t} - \frac{1}{t\left(1 + \frac{t}{2} + \frac{t^2}{6} + O(t^3)\right)}\\ &= \frac{1}{t} - \frac{1}{t}\left(1 - \left(\frac{t}{2} + \frac{t^2}{6} + O(t^3)\right) + \left(\frac{t}{2}+\frac{t^2}{6}+O(t^3)\right)^2 + O(t^3)\right)\\ &= \frac{1}{t} - \frac{1}{t}\left(1 - \frac{t}{2} + \frac{t^2}{12} + O(t^3)\right)\\ &= \frac{1}{2} - \frac{t}{12} + O(t^2), \end{align}$$

which is an entire meromorphic function with poles in $2\pi i k,\; k \in \mathbb{Z}\setminus\{0\}$ and nowhere else, the recursion $(1)$ shows that all $d_n$ are entire meromorphic functions with poles in $2\pi ik,\; k \in\mathbb{Z}\setminus\{0\}$ and nowhere else, and we have

$$d_{n+1}(0) = d_n(0) = d_0(0) = \frac{1}{2}$$

for all $n$. Thus

$$\sup_{t > 0} \left(\frac{1}{t} - f_n(t)\right) \geqslant \lim_{t\searrow 0} d_n(t) = \frac{1}{2},$$

and we see that the convergence is not uniform on any interval $(0,\varepsilon)$.

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