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I need to find the expression of function $f$, all we know about $f$ is:

$\begin{cases} \forall x>0, f(x+1)-f(x) = \frac{1}{x^2} \\ f \text{ is continuous on } ]0, +\infty[ \text{ and } \lim\limits_{x \to +\infty} f(x) = 0 \end{cases}$

Any help would be appreciated.

Emilio Novati
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anni
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  • Are you sure the problem is correct ? The result comes in terms of the first derivative of the digamma function. – Rebellos Nov 26 '18 at 21:19
  • It is correct, and it has one and only one solution. – anni Nov 26 '18 at 21:22
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    Yes, the solution is $f(x) = c_1 - ψ^{(1)}(x) + \frac{\pi^2}{6}$. But do you know what the digamma function and its derivative are ? – Rebellos Nov 26 '18 at 21:23
  • I just googled it, I have to say I am surprised because it seems to go past my calculus course, but since the problem has only one solution, if this one works it has to be the one. – anni Nov 26 '18 at 21:26
  • If it is well past your calculus course, then the exercise is mistaken. If I may ask, what kind of level of calculus course are you following ? – Rebellos Nov 26 '18 at 21:26
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    But maybe all the problem needs you to do is to show that $f(x)=-\sum\limits_{n=0}^\infty,\dfrac{1}{(x+n)^2}$ for all $x>0$, without knowing anything about the digamma function. It is not difficult to see that $f(x)$ is well defined (i.e., converges to a finite limit for all $x>0$). – Batominovski Nov 26 '18 at 21:32
  • I'm on my second year of math university. And the exercise wasn't mistaken, but I asked it a bit differently. The exercise just asked to show there was one and only one function satisfying those conditions, not actually find a solution. But I was curious to know the solution. – anni Nov 26 '18 at 21:33
  • @EulerPythagoras Well, then it makes a ton of difference. Try not to phrase exercises differently as you see what can happen. On the bright side of things, the existense of one and only solution can be shown by elaborationg Bataminovski's hint. – Rebellos Nov 26 '18 at 21:44

2 Answers2

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Since $f(t)\to 0$ as $t\to \infty$, we see that $$\sum_{n=0}^\infty\,\frac{1}{(x+n)^2}=\sum_{n=0}^\infty\,\big(f(x+n+1)-f(x)\big)=-f(x)\,.$$ This shows that the desired function $f:\mathbb{R}_{>0}\to\mathbb{R}$ must satisfy $$f(x)=-\sum_{n=0}^\infty\,\frac{1}{(x+n)^2}\tag{*}$$ for all $x>0$. The rest in the hidden portion is the justification that $f$ given by (*) satisfies the continuity requirement, as well as the limit requirement.

Note that the function given by (*) is well defined since the infinite sum converges by the comparison test with the series $\sum\limits_{n=1}^\infty\,\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$. The continuity of $f$ given in (*) follows from the observation that the sequence $\left\{f_k\right\}_{k\in\mathbb{Z}_{>0}}$ of functions $f_k:\mathbb{R}_{>0}\to\mathbb{R}$ defined by $$f_k(x):=-\sum_{n=0}^k\,\frac{1}{(x+n)^2}\text{ for all }k\in\mathbb{Z}_{\geq 0}\text{ and }x>0$$ uniformly converges to $f$. Clearly, we also have $$0>f(x)>-\frac{1}{x^2}-\sum_{n=1}^\infty\,\frac{1}{(x+n)(x+n-1)}=-\frac{1}{x^2}-\frac{1}{x}\,.$$ Ergo, $f$ indeed satisfies the limit requirement.

Batominovski
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I apologise for the following but couldn't unsee :

Since $f(x+1) - f(x) = 1/x^2$ and $f$ is defined over $\mathbb R^+$, then $x>0$ and this means that $x+1 > x$ and also $1/x^2 >0$. Thus $f(x+1) - f(x) > 0 \Leftrightarrow f(x+1) > f(x)$ and since that holds for every $x \in \mathbb R^+$, then $f$ is increasing and also $f$ is bounded, since $\lim_{x \to + \infty} f(x) = 0$.

Now, consider an operator $T$ such that $Tf(x) = f(x+1)$ defined as $T : C^b(0,+\infty) \to C^b(0,+\infty)$ where $C^b(0,+\infty)$ is the space of the continuous bounded functions in $(0,+\infty)$. In this space the sup norm is well defined and this space is complete (basically since the uniform limit of continuous functions is continuous). This tells us that $C^b(0,+\infty)$ is a Banach space.

Now, let $f,g \in C^b(0,+\infty)$ and $\lambda \in \mathbb R$. Then :

$$T(\lambda f + g) = (\lambda f + g)(x+1) = (\lambda f)(x+1) + g(x+1)$$

$$=$$ $$ \lambda f(x+1) + g(x+1) = \lambda Tf + Tg$$

This tells us that the operator $T$ is linear.

Now, it also is

$$\|Tf(x)\|_\infty = \|f(x+1)\|_\infty \leq \|f(x)\|_\infty$$

and thus $T$ is a bounded linear operator (specifically with $\|T\| \leq 1$) as well as $T \in B(C^b(0,+\infty))$.

But, since $C^b(0,+\infty)$ is Banach and $T \in B(C^b(0,+\infty))$, the equation

$$f(x) = \frac{1}{x^2} + Tf(x) \Leftrightarrow f(x) - Tf(x) = \frac{1}{x^2}$$

has a unique solution in $C^b(0,+\infty)$.

For the resemblance with the answer posted above, notice that

$$f(x) - Tf(x) = -\frac{1}{x^2} \Leftrightarrow (\mathbf{1} - T)f(x) = -\frac{1}{x^2} \Rightarrow f(x) = (\mathbf{1}-T)^{-1}\bigg(-\frac{1}{x^2}\bigg) $$

where $\mathbf{1}$ is the identity operator. But :

$$(\mathbf{1}-T)^{-1} = \sum_{n=1}^\infty T^n, \quad |T| < 1$$

Note : We defined $\frac{1}{x^2} : \mathbb R^+ \to \mathbb R$.

Note 2: Per comment discussion, if any doubts/issues regarding bounds, the space $C^b([\varepsilon, + \infty))$ could be considered where $\varepsilon >0$ and by manipulating $\varepsilon$ accordingly we could still yield all results.

Rebellos
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    Beautiful! Thanks for posting this. – Batominovski Nov 26 '18 at 22:22
  • @Batominovski Thanks for reading it ! I love that lemma, proving such staff is a treat ! – Rebellos Nov 26 '18 at 22:25
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    I like this a lot but I'm a little confused, since $x^{-2}$ isnt in $C^b$ do you need to use a weighted norm $\sup_{x>0} x^2 |f(x)| < \infty$? And shouldn't we need $|T| \underset{\color{red}\neq }{<} 1 $ to invert $\mathbf 1-T$? – Calvin Khor Nov 26 '18 at 22:31
  • @CalvinKhor Concerning the invert, he said "note the resemblance". He didn't mean that it's rigorous. – Scientifica Nov 26 '18 at 22:36
  • @Rebellos Beautiful answer! .But I think that Calvin Khor is right. In fact, $f$ is not bounded: looking at $f(x+1)-f(x)=\frac{1}{x^2}$, since $f$ is continuous, we see that $\lim_{x\to 0^+}f(x)=f(1)-\infty=-\infty$. – Scientifica Nov 26 '18 at 22:37
  • By the exercise, it shall be $x \in R^+$ and thus $x>0$, $\lim_{x \to \infty} f(x) = 0$ and $x \not\to 0$ which yields an essential bound idea. Regarding the invert, yes, resemblance is the only word but also $|T| \leq 1$ doesn't necessarily imply the equality. – Rebellos Nov 26 '18 at 22:38
  • @Scientifica Note that $x \in R^+$ thus it cannot be equal to zero, per the definition of the continuity of $f$ given in the excercise. $\text{ess}\inf$ would do a job. This is why I posted the last note in the approach ! – Rebellos Nov 26 '18 at 22:38
  • Also, I might be missing something, but having $f(x+1)>f(x)$ for all $x$ doesn't necessarily mean that $f$ is increasing. We might need to use more than this to show that $f$ is increasing. – Scientifica Nov 26 '18 at 22:39
  • @Scientifica Just let an arbitrary variable be expressed in terms of $x+1$ and it can be proven ! Use also that $\lim f(x) = 0$. – Rebellos Nov 26 '18 at 22:40
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    The boundedness of $x^{-2}$ can be easily fixed by consider $C^b\big([\epsilon,\infty)\big)$ instead and verify that $f|_{[\epsilon,\infty)}$ exists and is unique for each $\epsilon>0$. – Batominovski Nov 26 '18 at 22:44
  • @Batominovski Agreed! – Scientifica Nov 26 '18 at 22:44
  • @Batominovski +1 – Rebellos Nov 26 '18 at 22:47
  • @Batominovski i thought of that too but i liked the possiblity of the weight making $|T|$ strictly less than one (though I didn't check it) – Calvin Khor Nov 26 '18 at 22:47
  • @Rebellos Sure $x\neq 0$, but having $\lim_{x\to 0^+}f(x)=-\infty$ shows that it is not bounded. In fact, if you see the solution of the exercise as expressed in the answer above, you'll see that the solution $f$ is indeed unbounded. – Scientifica Nov 26 '18 at 22:47
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    @Scientifica Hmm, true in some way, but the essential idea of a bound could be yielded by Batominovski's comment and by manipulating $\varepsilon$ it can be extended. – Rebellos Nov 26 '18 at 22:51
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    Let $V$ be the subspace of functions with limit $0$ at $\infty$. Then, $S:=T|V$ sends $V$ to $V$. Note that, for each $f\in V$, $\lim{n\to \infty},S^nf=0$, so if $\sum_{n=0}^\infty,S^nf$ exists, then $$(1-S),\sum_{n=0}^\infty,S^nf=f,.$$ This shows that $\sum_{n=0}^\infty,S^nf$ is in the preimage of $f$ under $1-S$. – Batominovski Nov 26 '18 at 22:54
  • @Rebellos Agreed! As for the $f(x+1)>f(x)$ => $f$ increasing, here's a counter-example: take $\lfloor x\rfloor$. Sure, it is nondecreasing and not continuous. But you can modify it as follows: for each $n\in\mathbb Z$, make the function decrease a tiny bit and increase again around the middle point of $[n,n+1]$. You can also make the function continuous by connecting the discontinuities by taking a point like $(n+\frac{100}{101},n)$ and connecting it to the point $(n+1,n+1)$ in the graph of the floor function. The function you get from this transformation is not increasing yet $f(x+1)>f(x)$. – Scientifica Nov 26 '18 at 22:57
  • @Batominovski Nice! – Scientifica Nov 26 '18 at 22:57
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    In my previous comment, I proved existence, but not yet uniqueness. The uniqueness follows from the fact that the only function $f$ in $V$ such that $f=Sf$ is $f\equiv 0$. So, the issues are now resolved. – Batominovski Nov 26 '18 at 23:00
  • @Scientifica It's also $f : \mathbb R^+ \to R$ and $\lim_{x \to \infty} = 0$, which, as parameters, play a fact to the increasingness. – Rebellos Nov 26 '18 at 23:01
  • +1 for all of @Batominovski comments. I guess my approach yielded interest ! – Rebellos Nov 26 '18 at 23:02
  • @Rebellos I agree that all the other hypothesis matter. My point is that "$f$ is increasing" doesn't follow so easily from them. One could for example "prove it" by using the expression of the solution as given in the answer above. – Scientifica Nov 26 '18 at 23:09
  • Nevermind you didn't use "$f$ is increasing" anyway. – Scientifica Nov 26 '18 at 23:09
  • @Scientifica With the assumptions of the given problem it can be proven. I didn't elaborate on everything (for example that the space used is Banach). – Rebellos Nov 26 '18 at 23:21
  • @Rebellos It sure can be proven. I just said that how you put it doesn't make the proof clear. As I said in an earlier comment, I might be missing something, but again as I told you a bit later, one proof consists of taking the last result, namely $f(x)=-\sum\frac{1}{(x+n)^2}$, to show that it is increasing. That's a proof, isn't it? ;) – Scientifica Nov 26 '18 at 23:34
  • Sure ! Anyway, I didn't expect that my love for Banach spaces will attract such a nice conversation and attention! Thanks everyone for the discussion, the inputs and the justifications! – Rebellos Nov 26 '18 at 23:37
  • @Rebellos And I learned the technique from your answer. I also find amazing the similarity between $(I-T)^{-1}$ and the actual solution. Thank you for sharing your answer with us! – Scientifica Nov 26 '18 at 23:47
  • @Scientifica No problem, glad to share ! Well, my technique comes from the following lemma : Let X be a banach space and $T$ an operator $T \in B(X)$. If $| T | \leq \theta <1$, then for all $y \in X$ the equation $x = y + Tx$ has a unique solution. – Rebellos Nov 26 '18 at 23:51
  • Oh! But this doesn't seem to apply in your case as $|T|\le 1$ (in fact, $|T|=1$). – Scientifica Nov 26 '18 at 23:56
  • @Scientifica Extensions of this can be proven (there are adequate lemmas, alike the elaboration of Batominovski above, he proved exactly this). You can find similar lemmas involving the convergence of sequences which can be manipulated to give operators etc etc. The important thing is the Banach hypothesis! The general idea comes from it (which is the most strong of all lemmas). Weaker versions (like our elaborations) exist. – Rebellos Nov 27 '18 at 00:00
  • @Rebellos Ok. I'm curious to learn about these lemmas. Can you give me any reference please? – Scientifica Nov 27 '18 at 00:06
  • @Scientifica I am using Greek notes and stuff but I think Haim Breizis' book of functional analysis is very high regarded and of a very advanced level. – Rebellos Nov 27 '18 at 00:12
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    @Rebellos Ok thank you! I read parts of Brezis when I had a course on functional analysis, it's indeed an excellent book! – Scientifica Nov 27 '18 at 10:38
  • You need to, but did not, prove $f(x)$ is strictly increasing to show that $|T|<1$ to enable the Neumann series. – Hans Nov 29 '18 at 17:49