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I am looking for a non-trivial function $f(x)\in L_2(0,\infty)$ independent of the parameter $n$ (a natural number) satisfying the following integral equation: $$\displaystyle\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$$ or prove that $f(x)=0$ is the only solution.

The similar question is here but there are no parameters in the integral and the answer is based upon hit and trial method.

I want to know what would be the nice approach to tackle this problem.

EDIT: Such a function may exist, here is an example due to Stieltjes. A function $f(x) = \exp(-x^{1/4}) \sin x^{1/4}$ satisfies $\int_0^{\infty} f(x) x^n dx = 0$ for all integers $n \ge 0$.

We use the substitution $x=u^4$ to write $I_n = \int_0^{\infty} f(x) x^n dx = 4 \int_0^{\infty} e^{-u} \sin(u) u^{4n+3} du$; then integrate by parts four times (differentiating the power of $u$, and integrating the rest) to show that $I_n$ is proportional to $I_{n-1}$, and finally check that $I_0=0$.(the Edit copied from here)

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  • $\begingroup$ Does natural number include $n = 0$? (Not that I seriously expect that to affect the answer; I'm just curious.) $\endgroup$ – Brian Tung Apr 22 at 16:49
  • $\begingroup$ @BrianTung, NO, $n=1,2,3...$ $\endgroup$ – ersh Apr 22 at 17:11
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    $\begingroup$ I've been attacking this since yesterday. What I did was a change of variable $y=e^x$ to get the integral $$\int_1^{\infty} \frac{f(\log(y))}{1+y^n} \frac{dy}{y}.$$ Defining $g(y) = f(\log(y)) \chi_{[1,\infty)}(y)$, the integral is $$\int_0^{\infty} \frac{g(y)}{1+y^n} \frac{dy}{y}$$ which looks like a Haar integral. $\frac{dy}{y}$ is the Haar measure for the group $G = (0,\infty)$ with multiplication. If you use Pontryagin duality and isometry of the Fourier transform, you could turn this into a new inner product. I'm not sure how useful this is though. $\endgroup$ – Cameron Williams Apr 25 at 15:52
  • $\begingroup$ Why do you award the bounty to an answer that you didn’t accept? I think the reasoning is “I will accept fedja’s answer because others believe it is the right answer to the problem, but I will award the bounty to jawheele’s answer, which gets me the furthest in that direction in ways that I understand.” $\endgroup$ – user210229 Apr 29 at 12:25
  • $\begingroup$ @Matt F., I am convinced by fedja's proof though I am still learning the background details. The bounty got distributed itself to other answer. I didn't want to give it by myself since other answer is not complete. I already had made the intention yo accept fedja's answer even though I was still figuring out certain background results which he used. $\endgroup$ – ersh Apr 29 at 13:54
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The only such function is $f\equiv 0$. The proof is fairly standard but if you've never seen such machinery before, it may take some time and effort to comprehend, so feel free to ask questions if something is unclear.

Lemma: If $f$ satisfies the condition, then $F(z)=\int_{0}^\infty \frac{f(x)}{1+e^{zx}}dx=0$ for all real $z>0$. Proof: $F$ is analytic in the right half-plane $\Re z>0$ and, moreover, grows at most polynomially in $\Re z>1$, say. Thus $z^{-N}F(z)$ is a bounded analytic function in $\Re z>1$. By hypothesis, $F$ has zeroes at all integer points, and $F$ must vanish identically (the Blaschke condition is violated). This proves the lemma.

By the lemma, if $f$ satisfies the condition, then $\int f(x)K_t(x)dx=0$ for all $t$, where $K_t(x)=\frac 1{1+e^{tx}}$. So the main result will follow from the density of the family $K_t$ in $L^2(0,+\infty)$. This the same as the density of the family $k_t$ in $L^2((0,+\infty),\frac{dx}{x})$ where $k_t(x)=\frac{\sqrt{tx}}{1+e^{tx}}$. Up to a logarithmic change of variable, this is also the same as the problem about completeness of shifts in $L^2(\mathbb R)$. The answer is given by the Wiener criterion: the family is dense if the Fourier transform of the generating function vanishes only on a set of measure $0$. So the family is dense if the transform $$ \int_0^\infty \frac{\sqrt x}{1+e^x}x^{-is}\,\frac{dx}x\ne 0 $$ for almost all $s\in\mathbb R$. But the left hand side is analytic in $s$ for $|\Im s|<\frac 12$, and not identically zero, so the real zeroes of the transform are at most a discrete set, and the transforms are non-zero for almost all $s$. QED

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  • $\begingroup$ Will you give a reference for this use of the Wiener criterion? I found a surprisingly appropriate reference for the main result used in the previous paragraph. $\endgroup$ – user210229 Apr 26 at 20:26
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    $\begingroup$ @MattF. Oh, my. Finding references is my weak point. You can always cite Wikipedia (en.wikipedia.org/wiki/Wiener%27s_tauberian_theorem), of course, but the article there does not contain a proof. On the other hand, the proof can be found in almost every book on harmonic analysis, but it may take some time to find an exposition freely and legally available online... $\endgroup$ – fedja Apr 26 at 22:30
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    $\begingroup$ @ersh Regarding (1), the result follows from the density of the family $K_t$ in $L^2(0,\infty)$ because $\int fK_t = 0 \iff \langle f , K_1 \rangle = 0$, where $\langle \cdot,\cdot \rangle$ denotes the $L^2$ inner product, so if the family is dense, then continuity of the inner product implies $\langle f,g \rangle = 0$ $\forall g \in L^2(0,\infty)$, so positive definiteness of the inner product implies $f=0$ a.e.. $\endgroup$ – jawheele Apr 27 at 2:54
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    $\begingroup$ @ersh Yes, $\frac{dx}x$ is the measure with respect to which $L^2$ is taken. As to the first equivalence, just note that $\|g(x)-\sum_t c_tK(tx)\|_{L^2(dx)}=\|g(x)\sqrt x-\sum_t c_t t^{-1/2}k(tx)\|_{L^2(\frac{dx}{x})}$. The last equivalence is just the isometry $f(x)\mapsto f(e^y)$ between $L^2((0,\infty),\frac{dx}{x})$ and $L^2(\mathbb R,dy)$ and the observation that the logarithm of the product is the sum of the logarithms. $\endgroup$ – fedja Apr 27 at 17:57
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    $\begingroup$ @ersh No need for the authorship: the argument is really quite standard, so my contribution reduces to just writing it down. If you want to acknowledge it in some way, just cite MO :-) $\endgroup$ – fedja Apr 28 at 13:33
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I'll prove a partial result that hopefully someone can complete to showing that $f(x) \equiv 0$ is the unique continuous and $L^2(\mathbb{R}_{>0})$ solution. Beyond that, perhaps one can use that continuous functions are dense in $L^2$ to get the requested result. The main tool I want to contribute is the following lemma:

$\mathbf{Lemma}:$ Consider $f \in C(\mathbb{R}_{>0}) \cap L^2(\mathbb{R}_{>0})$. Then for any choice of $0<x_0<y_0 \leq \infty$, we have

$$f(x_0) = \lim_{n \to \infty} ne^{nx_0}\int_{x_0}^{y_0}\frac{f(x)}{1+e^{nx}}dx $$

$\mathbf{Proof}:$ Fix $\epsilon>0$ and, by continuity, choose $\delta>0$ such that $|x-x_0| < \delta \implies|f(x)-f(x_0)| < \epsilon$. Note that

$$\lim_{n \to \infty} ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{1}{1+e^{nx}}dx = 1$$

and

\begin{align} \lim_{n \to \infty} \left | n e^{nx_0} \int_{x_0+\delta}^{y_0}\frac{f(x)}{1+e^{nx}} dx \right | & \leq \lim_{n \to \infty} n e^{nx_0} \int_{x_0+\delta}^{y_0} e^{-nx} |f(x)| dx \\ & \leq \lim_{n \to \infty} n e^{nx_0} e^{-(n-1)(x_0+\delta)} \int_{x_0+\delta}^{y_0} e^{-x} |f(x)| dx \\ & \leq \lim_{n \to \infty} n e^{x_0+\delta} e^{-n\delta} \|e^{-x}f(x)\|_1 \\ & \leq \lim_{n \to \infty} n e^{x_0+\delta} e^{-n\delta} \|e^{-x}\|_2 \cdot \|f\|_2 \\ & = 0 \end{align}

by Holder's inequality (neither of these depended on the choice of $\delta$-- they're just straightforward computation/comparison). Given these results, we see that

\begin{align} \lim_{n \to \infty} \left |ne^{nx_0} \int_{x_0}^{y_0} \frac{f(x)}{1+e^{nx}}dx - f(x_0) \right | & = \lim_{n \to \infty} \left |ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{f(x)-f(x_0)}{1+e^{nx}}dx \right | \\ & \leq \lim_{n \to \infty} ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{|f(x)-f(x_0)|}{1+e^{nx}}dx \\ & \leq \epsilon \end{align}

Since $\epsilon>0$ was arbitrary, this proves the lemma. $$\tag*{$\blacksquare$}$$

This lemma gives us some insight into the behavior of functions satisfying your condition. In particular, if $f$ satisfies these hypotheses and the condition under consideration, we must have for any $x_0 > 0$ that

$$0 = \lim_{n \to \infty} ne^{nx_0} \int_{0}^\infty \frac{f(x)}{1+e^{nx}}dx = f(x_0) + \lim_{n \to \infty} ne^{nx_0} \int_{0}^{x_0} \frac{f(x)}{1+e^{nx}}dx$$ Or

$$f(x_0) = -\lim_{n \to \infty} ne^{nx_0} \int_0^{x_0} \frac{f(x)}{1+e^{nx}}dx $$

From this result, we have $\forall$ $0<x_0<y_0<\infty$ that

$$\lim_{n \to \infty} ne^{nx_0} \int_0^{y_0} \frac{f(x)}{1+e^{nx}}dx = \lim_{n \to \infty} e^{-n(y_0-x_0)} \left [ ne^{ny_0} \int_0^{y_0} \frac{f(x)}{1+e^{nx}}dx \right ] = 0 \cdot (-f(y_0)) = 0 $$

and similarly $\forall$ $0 < w_0 < x_0$ with $f(w_0) \neq 0$, the expression

$$ne^{nx_0} \int_0^{w_0} \frac{f(x)}{1+e^{nx}}dx = e^{n(x_0-w_0)} \left [ ne^{nw_0} \int_0^{w_0} \frac{f(x)}{1+e^{nx}}dx \right ] $$ diverges to $-\text{sgn}(f(w_0)) \cdot \infty$ as $n \to \infty$, since the term in brackets approaches $-f(w_0)$.

Now, suppose $f$ is not identically $0$, so the set

$$\{t >0 | f \equiv 0 \text{ on } (0,t) \}$$

is bounded above; define $T_1$ as the supremum of this set if it is nonempty and $0$ if it is empty. Suppose $\exists T_2>T_1$ such that $$ T_1 < t < T_2 \implies f(t) \neq 0$$ Then $f$ is nonzero on $(T_1,T_2)$ by definition, so by the Intermediate Value Theorem $f$ cannot change sign in this interval. Thus either $f(x) \geq 0$ or $-f(x) \geq 0$ on $(0,T_2)$-- WLOG assume the former, so that for any $x_0 \in (T_1,T_2)$ the first identity after the lemma gives that $f(x_0) \leq 0$, showing $f(x_0)=0$, a contradiction. We therefore conclude that no such $T_2$ exists, i.e. $\forall t>T_1$ $\exists s \in (T_1,t)$ such that $f(s)=0$. That is to say, on any interval containing $0$ on which $f$ is not identically zero, $f$ must have infinitely many roots greater than $T_1$. In particular, the suggestion $f(x)=e^{-x^{1/4}} \sin(x^{1/4})$ cannot be a solution.

Note that if one can use the above results to show that $T_1 = 0$, it would establish that no continuous solution with compact support on $\mathbb{R}_{>0}$ exists (relevant because such functions are dense in $L^2(\mathbb{R}_{>0})$).

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  • $\begingroup$ How does "$T_1=0$" imply the result? And I guess you mean "on any interval containing $T_1$" there. $\endgroup$ – XIAODA QU Apr 24 at 4:21
  • $\begingroup$ @XIAODAQU The statement $T_1=0$ is exactly saying that the support of $f$ contains points arbitrarily close to the boundary, so the support is not compact in $(0,\infty)$. Also, any interval containing $0$ in which $f$ is not identically zero contains $T_1$, but I agree that "containing $T_1$" would suffice-- I said $0$ to help with the intuitive interpretation of the statement. $\endgroup$ – jawheele Apr 24 at 4:27
  • $\begingroup$ If it's useful to anyone, the hypotheses of the lemma can be relaxed to $f$ continuous and $f(x)e^{-x} \in L^1((0,\infty))$ (so, for example, $f$ continuous and bounded by a polynomial would suffice, etc.). $\endgroup$ – jawheele Apr 24 at 4:48
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    $\begingroup$ @jawheele , I will read your solution thoroughly tomorrow. I have possibly constructed a non-zero solution which I am gonna post tomorrow. Thanks for your time! $\endgroup$ – ersh Apr 24 at 4:53
  • $\begingroup$ Sorry for late reply. The solution I had constructed works only if the domain contains origin, specifically on $L_2(-1,\infty )$. I tried to fix it but unsuccessful. One possible way to make it work is to make a change in the Fourier transform (by replacing $e^{-2\pi i x t}$ by $e^{-2\pi i x (t-4)}$. So this would be a new transform and it is lot of work to study it. $\endgroup$ – ersh Apr 27 at 1:05
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Edit: Assuming a non-zero $f(x)$ exists - if it is antisymmetric about $x=0$ and can be expressed by a Fourier integral $$f(x) = \int_{-\infty}^\infty F(\omega) \sin(\omega x) {\rm d}\omega ,$$ it can be shown that $$\lim_{n\rightarrow\infty}\int_0^\infty \frac{f(x)}{1+\exp(nx)}{\rm d}x = 0 :$$

for all spectral components of the assumed antisymmetric $f(x)$ proportional to $\sin(\omega x)$, the integral $$\int_0^\infty \frac{\sin(\omega x)}{1+\exp(nx)}{\rm d}x = \frac{n\sinh(\frac{\pi\omega}{n})-\pi\omega}{2n\omega\sinh(\frac{\pi\omega}{n})}$$ vanishes for $n\rightarrow\infty$ (and also for $\omega\rightarrow\infty$ since $n$ is non-zero).

And here the previous attempt at a numerical solution...
The approximate $f(x)$ in the plot below was constructed by numerically integrating $f(x)/(1+\exp(nx))$ and considering all $n$ up till including $30000$ and sampling $f(x)$ on a logarithmic-like grid with $15000$ points. $f(x)$ appears to consist of a superposition of functions of the form $\sin({\rm const} + \log({\rm const}_1+{\rm const}_2\cdot{}x))$, but it's hard to deduce a closed form...

Approximate f(x) constructed numerically.

The fact that this numerical solution stops oscillating at $x\sim10$ likely is an artifact of limited numerical precision; due to $(1+\exp(nx))^{-1}$ in the kernel, there is very little weight on $f(x)$ at larger $x$ and it is hard to obtain more precise values for $f(x)$.
For large $n$, which should have a dominating influence on $f(x)$ for $x\rightarrow 0$, the finite number of grid points limits the qualitity of the solution for $f(x)$. For sufficiently large $n$ and with a finite number of grid points, $f(x\rightarrow 0)\rightarrow 0$ will be sufficient to satisfy the numerical solution attempted here.

Here is the Python code used to generate the figure above:

#number of grid points
N = 15000
#max n considered (can be/should be larger than N for least-squares
#solution below)
n = 2*N

#external dependencies
import numpy as np
from numpy.polynomial.legendre import leggauss

#create quadrature roots and weights
#beginning with interval [-1;1]
xp,wp = leggauss(N)
xp *= 0.5
xp += 0.5
#rescale to 0..infinity
scale = 6.0
x = np.sinh(scale*xp)
w = wp * scale/2. * np.cosh(scale*xp)

#create system of equations: Integral(n) = 0, n=0...
#we also constrain Int(0) = 0 (although only n=1... required) ...
#... seems to be more stable numerically (otherwise almost no weight on
#f-values at large x)
A = []
for i in range(n+1):
    A.append(np.empty_like(x))
    for j in range(len(x)):
        if x[j]*i<500.:
            A[-1][j] = w[j]/(1.+np.exp(x[j]*i))
        else:
            A[-1][j] = 0.0

#add additional for f at x approx 0.5 to avoid trivial solution f=0
#(hoping to not accidentally hit a root of f(x))
z = np.zeros_like(x)
z[np.argmin(np.abs(x-0.5))] = 1.
A.append(w)

A = np.array(A)

#RHS of system of equations
b = np.concatenate((np.zeros(n+1,np.float),[1.]))

#least-squares solution for f(x)
sol = np.linalg.lstsq(A,b,rcond=1e-14)[0]
#normalize
if -np.min(sol)>np.max(sol):
    sol *= -1.0
sol /= np.max(sol)

#plot solution
import pylab
pylab.plot(x,sol)
pylab.xscale('log')
pylab.xlabel('$x$')
pylab.ylabel('$f(x)$')
pylab.show()
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  • $\begingroup$ I really appreciate your numerical attempt. Thanks for your time! $\endgroup$ – ersh Apr 24 at 4:55
  • $\begingroup$ Just goes to show - if it's a problem that "seems very strongly" like it should have no solution, and yet it does, that solution's gonna likely be WEEEEEEERD! $\endgroup$ – The_Sympathizer Apr 26 at 10:06
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The linearized version has a nice solution. If we use a Taylor series at $x=0$ to approximate $$\frac{1}{1+e^{nx}}\ \simeq\ \max\left(\frac{1}{2} - \frac{nx}{4},\ 0\right)$$ then the linearized problem asks for a non-trivial $f$ in $L^2$ with $$\int_0^{2/n}\left(\frac{1}{2} - \frac{nx}{4}\right)f(x)dx = 0$$ valid for all $n$.

Let $$ f_m(x)= \begin{cases} +1\ \ \text{ if }\ \ x \in (\frac2{m+1}, \phantom{+ 3a_m},\ \frac2{m+1} + \phantom{3}a_m)\\ -1\ \ \text{ if }\ \ x \in (\frac2{m+1} + \phantom{3}a_m, \ \frac2{m+1} + 3a_m),\text{ where }a_m=\dfrac{1}{2m(m+1)}\\ +1\ \ \text{ if }\ \ x \in (\frac2{m+1} + 3a_m, \ \frac2{m+1} + 4a_m)\\ \ \ 0 \ \ \ \ \text{otherwise.} \end{cases} $$ $\ \ $f_m(x) for m=6

Each $f_m$ is non-zero on the interval $(2/(m+1), 2/m)$. Furthermore the product of $f_m$ with any linear function has an integral of $0$, because the middle half of a trapezoid has the same area as the two outer quarters.

(1/2-x)f_m(x) for m=6

So $$f=\sum_{m=1}^\infty f_m$$ is an $L^2$ function, alternating between $+1$ and $-1$, with a square-integral of $2$, which solves the linearized problem.

For the original problem, we could try a similar technique by dividing the interval $(2/(m+1), 2/m)$ into $m$ different subintervals so that all of the relevant integrals are $0$.

Alternatively, any proof that the original problem has no solution will have to use some surprisingly detailed properties of $1/(1+e^{nx})$.

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The issue condition can be written in the form of $$\int_0^\infty f\left(\dfrac zn\right)\dfrac{dz}{1+e^z} = 0.\tag1$$ Let the Maclaurin series of $f(x)$ exists, then $$f(z) = \sum\limits_{k=0}^\infty a_kz^k.$$

Taking in account the integral representation of the Riemann zeta function in the form of

$$\int\limits_0^\infty\dfrac {z^k\,dz}{1+e^z} = (1-2^{-k})k!\zeta(k+1)\tag3$$

(see also Wolfram Alpha),

one can get $$I_n = \sum\limits_{k=0}^\infty \dfrac{a_k}{n^k} \int\limits_0^\infty\dfrac {z^k\,dz}{1+e^z} = \sum\limits_{k=0}^\infty \dfrac{1-2^{-k}}{n^k}k!\zeta(k+1) = 0,$$ $$I_n = \sum\limits_{k=0}^\infty \dfrac{b_k}{n^k} = 0\quad \forall (n\in \mathbb N),\tag4$$ where $$b_k = (1-2^{-k})k!\zeta(k+1).\tag5$$ The system $(4)$ is linear relatively to the coefficients $b_k.$ On the other hand, determinant of this system is

$$ \Delta = \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\ 1 & \dfrac12 & \dfrac1{2^2} & \dfrac1{2^3} & \dfrac1{2^4} & \dots & \dfrac1{2^k} &\dots\\ 1 & \dfrac13 & \dfrac1{3^2} & \dfrac1{3^3} & \dfrac1{3^4} & \dots & \dfrac1{3^k} &\dots\\ 1 & \dfrac14 & \dfrac1{4^2} & \dfrac1{4^3} & \dfrac1{4^4} & \dots & \dfrac1{4^k} &\dots\\ &&&&\dots\\ 1 & \dfrac1n & \dfrac1{n^2} & \dfrac1{n^3} & \dfrac1{n^4} & \dots & \dfrac1{n^k} &\dots\\ \end{vmatrix}\\[6pt] = \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\ 0 & -\dfrac12 & -\dfrac1{2^2} & -\dfrac1{2^3} & -\dfrac1{2^4} & \dots & -\dfrac1{2^k} &\dots\\ 1 & -\dfrac23 & -\dfrac2{3^2} & \dfrac2{3^3} & \dfrac2{3^4} & \dots & -\dfrac2{3^k} &\dots\\ 1 & -\dfrac34 & \dfrac3{4^2} & -\dfrac3{4^3} & -\dfrac3{4^4} & \dots & -\dfrac3{4^k} &\dots\\ &&&&\dots\\ 1 & -\dfrac{n-1}n & -\dfrac{n-1}{n^2} & -\dfrac{n-1}{n^3} & -\dfrac{n-1}{n^4} & \dots & \dfrac1{n^k} &\dots\\ \end{vmatrix}\\[6pt] \sim \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\ 0 & 1 & \dfrac12 & \dfrac1{2^2} & \dfrac1{2^3} & \dots & \dfrac1{2^k} &\dots\\ 0 & 1 & \dfrac13 & \dfrac1{3^2} & \dfrac1{3^3} & \dots & \dfrac1{3^k} &\dots\\ 0 & 1 & \dfrac14 & \dfrac1{4^2} & \dfrac1{4^3} & \dots & \dfrac1{4^k} &\dots\\ &&&&\dots\\ 0 & 1 & \dfrac1n & \dfrac1{n^2} & \dfrac1{n^3} & \dots & \dfrac1{n^k} &\dots\\ \end{vmatrix} \not = 0.$$

If the homogenius linear system has unzero determinant, then zero solution is single.

Looks that consideration of the determinant sequences allows to assume that $\textbf{the issue task has only the trivial solution}$.

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  • $\begingroup$ @ Is term by term integration justified here? $\endgroup$ – ersh Apr 27 at 20:31
  • $\begingroup$ Also, making that substitution, makes $f$ so $a_n$ dependent on parameter $n$? $\endgroup$ – ersh Apr 27 at 20:37
  • $\begingroup$ @ersh Thanks for the comments. 1) The integral representation of zeta is the basic property (link is added to the answer). 2) The substitution is not very principial thing, another ways lead to the same result. $\endgroup$ – Yuri Negometyanov Apr 27 at 21:04
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    $\begingroup$ As ersh pointed out, swapping the limits doesn't seem justified in general, and saying $f$ equals its Maclaurin series already requires that $f$ is analytic on $\mathbb{R}$. This is quite a long way from saying $f \equiv 0$ is the unique $L^2(0,\infty)$ solution. $\endgroup$ – jawheele Apr 27 at 21:37
  • $\begingroup$ @jawheele Thanks for the comment. Swapping the limits shows the real complexity of the task. I see $f(x)$ quite good function to satisfy very hard conditions. $\endgroup$ – Yuri Negometyanov Apr 27 at 22:00
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There is no one continuous function that will satisfy that for all n.

By trivial solution I assume you mean $f(x)=0$. One potential (although also fairly trivial, or at least boring) answer is

$$f(x)=(1+e^{nx})g(x)$$

where

$$\int_0^{\infty} g(x) \, dx=0$$

say maybe $\sin(x)$ or indeed any periodic function that acts in this way. However, this of course is not strictly defined as our $\epsilon-\delta$ definition will not hold for convergence, i.e.

$$\lim_{x\to\infty} \int_0^n \sin(x) \, dx$$

is not defined, so that is not an option either.

Let us now consider to look at this graphically. Imagine, if you can, the graph of such a function, say defined from $0$ to $\infty$ such that our integral is zero. Without a loss of generality, let us say that this function is positive to start with, so at some point, say $x_0$ this function will have to cross the $x$ axis and become negative.

This means that our function has to be periodic and increasing at exactly the same rate and time as $1+e^{nx}$, thus resulting in only the trivial solution as above, or that the finitely bounded area before this critical value $x_0$ has identical area to our unbounded area beyond $x_0$.

This means that our function $f(x)$ satisfies $$\int_0^{x_0}f(x) \, dx=-\int_{x_0}^\infty f(x) \, dx$$

As the function must satisfy this for all n this answer should suffice to show that the function $f(x)$ must be trivial, as if $f(x)$ satisfies this for some n, the the denominator will grow at a different rate for a different n, thus showing that there is no such $f(x)$.

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    $\begingroup$ The question specified the solution should solve the equation for all $n>0$. $\endgroup$ – jawheele Apr 22 at 19:36
  • $\begingroup$ Have edited the answer to reflect that, thank you $\endgroup$ – W M Seath Apr 22 at 19:42
  • $\begingroup$ You haven't proven that no continuous function exists for all $n$, which is what the question specifically asks for-- you've just claimed it. $\endgroup$ – jawheele Apr 22 at 19:43
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    $\begingroup$ What stops $f$ oscillating wildly? See the Stieltjes example above where all the momentums are zero. Something similar may work here too or not but it needs proof not hand waving $\endgroup$ – Conrad Apr 22 at 19:45
  • $\begingroup$ Your f(x) solution is interesting, but the problem states that it should independent from $n$. $\endgroup$ – The Count Apr 24 at 22:45

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