Jtdylktuy

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

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

Any help would be appreciated.

Since $f(t)to 0$ as $tto infty$, 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 $f:mathbb{R}_{>0}tomathbb{R}$ 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 $sumlimits_{n=1}^infty,dfrac{1}{n^2}=dfrac{pi^2}{6}$. The continuity of $f$ given in (*) follows from the observation that the sequence $left{f_kright}_{kinmathbb{Z}_{>0}}$ of functions $f_k:mathbb{R}_{>0}tomathbb{R}$ defined by $$f_k(x):=-sum_{n=0}^k,frac{1}{(x+n)^2}text{ for all }kinmathbb{Z}_{geq 0}text{ and }x>0$$ 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.

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.