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I am working on understanding the resolvent set of a certain eigenvalue problem. The solutions of the resolvent equation involve integrals similar to what I present below. Proving or contradicting the result claimed below would be useful.

Consider the expression below defined for $x\geq 0$ as: $$ g(x)=x^p\int_1^x\left(\frac{f(x')}{x'^{p+1}}\right)dx'\text{ where }p>0. $$ If $f$ is analytic, then it easy to prove (by expanding $f$ and integrating term-by-term) that $$ g(x)=-\frac{f(x)}{p}+\mathcal{O}(x)+\mathcal{O}(x^p)\text{ as }x\to 0^+.\tag{1} $$ On the RHS of (1), I really mean "$f(x)$", not "$f(0)$", because, below, I will allow the case where $f(0)$ is not necessarily defined.

I really would like to extend (1) to cases where $f$ is "less nice", that is:

(1) If $f$ is absolutely continuous everywhere.

(2) If $f$ is continuous and bounded on $(0,\infty)$. If that helps, also in $H^1(a,\infty)$ for any $a>0$.

Using L'Hopital's rule, it is not difficult to prove the following for the cases (1) and (2) above: $$ g(x)=-\frac{f(x)}{p}+\mathcal{o}(1).\tag{2} $$ Actually, if, in (2), I could replace "$\mathcal{o}(1)$" by "$\mathcal{O}(x^\epsilon)$" for some $\epsilon>0$, I would be happy, even if (1) would not be extended.

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2 Answers 2

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Taking in account formulas for the derivative of production, one can get: $$\text I(x)=\left(\dfrac {x^{p+1}}{p+1}\right)'_x\ \cdot\left(\int\limits_1^x \dfrac1{t^{p+1}}\,f(t)\,\text dt\right)$$ $$=\left(\dfrac {x^{p+1}}{p+1} \cdot \int\limits_1^x \dfrac1{t^{p+1}}\,f(t)\,\text dt\right)'_x -\left(\dfrac {x^{p+1}}{p+1}\right)\,\left(\int\limits_1^x \dfrac1{t^{p+1}}\,f(t)\,\text dt\right)' $$ $$=\left(\dfrac{x}{p+1}\,\text I(x)\right)' -\left(\dfrac {x^{p+1}}{p+1}\right)\,\dfrac1{x^{p+1}}\,f(x) =\left(\dfrac{x}{p+1}\text I(x)\right)'-\dfrac {f(x)}{p+1},$$ $$(p+1)\,\text I(x) =x\,\text I'(x)+\text I(x) -f(x),$$ $$x\,\text I'(x)-p\,\text I(x)=f(x),\quad I(1)=0.\tag1$$ Easily to see that the equation $(1)$ in the form of $$x^{p+1}\left(x^{-p}\text I\right)' =f(x)$$ corresponds to the given task.

Let $$f(x)=\sum\limits_{j=0}^\infty a_j\,(x-1)^j,\quad \text I(x)=\sum\limits_{k=1}^\infty b_j\,(x-1)^j,\,\tag2$$ then $$\sum\limits_{j=0}^\infty a_j\,(x-1)^j =\sum\limits_{j=1}^\infty jb_j\,(x-1)^j +\sum\limits_{j=1}^\infty b_j\,(x-1)^{j-1} -p\sum\limits_{j=1}^\infty b_j\,(x-1)^j$$ $$=b_1+\sum\limits_{j=1}^\infty \big((j-p)b_j+b_{j+1}\big)\,(x-1)^j,$$

\begin{cases} b_1=a_0\\ b_2=a_1+(p-1)b_1\\ b_3=a_2+(p-2)b_2\dots\\ b_{j+1}=a_{j}+(p-j)b_j\dots\tag3 \end{cases}

Recurrence relations $(3)$ leads to the fast increment of $b_j.$ However, convergence can be achieved for the parameters $\;p=5, a_j=\dfrac1{(5j)!};$

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    $\begingroup$ Hey thank you very much. The issue is that $f$ is continuous only, and thus cannot be expanded in series like you do. However, the idea of writing this differential equation for $I$ might be very useful. $\endgroup$ Nov 7, 2022 at 22:02
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$\def\d{\mathrm{d}}\DeclareMathOperator{\sgn}{sgn}\def\abs#1{\left|#1\right|}\def\bracket#1{\left[#1\right]}\def\paren#1{\left(#1\right)}\def\peq{\mathrel{\phantom=}}$For Condition 2:

$f$ is continuous and bounded on $(0, +∞)$, and $H^1$ on $(a, +∞)$ for any $a > 0$,

note that the ultimate task only cares about the values of $f$ near $0$, so the $+∞$ above can be modified to be $1$, and now there is a counterexample to this condition: $f(x) = \sin\dfrac{1}{x^p}$.

Since$$ \int_1^x \frac{f(u)}{u^{p + 1}} \,\d u = \int_1^x \frac{1}{u^{p + 1}} \sin\frac{1}{u^p} \,\d u = -\frac{1}{p} \cos\frac{1}{u^p}\left.\vphantom\intop\right|_1^x = \frac{1}{p} \paren{ \cos 1 - \cos\frac{1}{x^p} }, $$ then$$ g(x) + \frac{1}{p} f(x) = \frac{x^p}{p} \paren{ \cos 1 - \cos\frac{1}{x^p} } + \frac{1}{p} \sin\frac{1}{x^p} = \frac{1}{p} \sin\frac{1}{x^p} + O(x^p) \ (x → 0^+). $$ The term $\dfrac{1}{p} \sin\dfrac{1}{x^p}$ is not even $o(1)$ as $x → 0^+$.


For Condition 1:

$f$ is absolutely continuous on $(0, +∞)$,

there is also a counterexample:\begin{gather*} h(x) = x \sum_{n = 2}^∞ \frac{2^{3n}}{n^2} (2^{-2n} - |x - 2^{-n}|) I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x),\\ f(x) = h(x) - p \int_1^x \frac{h(u)}{u} \,\d u\ (x > 0), \end{gather*} where $I$ is the indicator function.

Proposition 1: $h$ is bounded on $(0, +∞)$.

Proof: Note that the intervals $[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]$ are mutually disjoint for $n \geqslant 2$. For $x > 0$,\begin{align*} 0 \leqslant h(x) &\leqslant \sum_{n = 2}^∞ x · \frac{2^{3n}}{n^2} · 2^{-2n} I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x)\\ &\leqslant \sum_{n = 2}^∞ (2^{-n} + 2^{-2n}) · \frac{2^n}{n^2} I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x)\\ &= \sum_{n = 2}^∞ \frac{1}{n^2} (1 + 2^{-n}) I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x)\\ &\leqslant \sup_{n \geqslant 2} \frac{1}{n^2} (1 + 2^{-n}) < +∞. \tag*{$\square$} \end{align*}

Proposition 2: $h$ is absolutely continuous on $(0, +∞)$.

Proof: For $x > 0$,\begin{align*} h'(x) &= \sum_{n = 2}^∞ \frac{2^{3n}}{n^2} (x (2^{-2n} - |x - 2^{-n}|))' I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x)\\ &= \sum_{n = 2}^∞ \frac{2^{3n}}{n^2} (2^{-2n} - |x - 2^{-n}| - x\sgn(x - 2^{-n})) I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x), \end{align*} and it suffices to prove that $h' \in L^1((0, +∞))$.

For $n \geqslant 2$, if $x \in [2^{-n} - 2^{-2n}, 2^{-n}]$, then\begin{gather*} 2^{-2n} - |x - 2^{-n}| - x\sgn(x - 2^{-n}) = 2x - 2^{-n} + 2^{-2n} > 0; \end{gather*} if $x \in [2^{-n}, 2^{-n} + 2^{-2n}]$, then$$ 2^{-2n} - |x - 2^{-n}| - x\sgn(x - 2^{-n}) = -2x + 2^{-n} + 2^{-2n} < 0. $$ Therefore$$ |h'(x)| = \sum_{n = 2}^∞ \frac{2^{3n}}{n^2} |2^{-2n} - |x - 2^{-n}| - x\sgn(x - 2^{-n})| I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x), $$ and\begin{align*} &\peq \int_{(0, +∞)} |h'(x)| \,\d x\\ &= \sum_{n = 2}^∞ \int_{(0, +∞)} \frac{2^{3n}}{n^2} |2^{-2n} - |x - 2^{-n}| - x\sgn(x - 2^{-n})| I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x) \,\d x\\ &= \sum_{n = 2}^∞ \frac{2^{3n}}{n^2} \paren{ \int_{2^{-n} - 2^{-2n}}^{2^{-n}} (2x - 2^{-n} + 2^{-2n}) \,\d x + \int_{2^{-n}}^{2^{-n} + 2^{-2n}} (2x - 2^{-n} - 2^{-2n}) \,\d x }\\ &= \sum_{n = 2}^∞ \frac{2^{3n}}{n^2} · 2^{-3n + 1} = \sum_{n = 2}^∞ \frac{2}{n^2} < +∞. \tag*{$\square$} \end{align*}

Propodision 3: $\dfrac{h(x)}{x} \in L^1((0, +∞))$.

Proof:\begin{align*} &\peq \int_{(0, +∞)} \abs{ \frac{h(x)}{x} } \,\d x\\ &= \sum_{n = 2}^∞ \int_{(0, +∞)} \frac{2^{3n}}{n^2} (2^{-2n} - |x - 2^{-n}|) I_{[2^{-n} - 2^{-2n},\, 2^{-n} + 2^{-2n}]}(x) \,\d x\\ &= \sum_{n = 2}^∞ \frac{2^{3n}}{n^2} \paren{ \int_{2^{-n} - 2^{-2n}}^{2^{-n}} (x - 2^{-n} + 2^{-2n}) \,\d x + \int_{2^{-n}}^{2^{-n} + 2^{-2n}} (-x + 2^{-n} + 2^{-2n}) \,\d x }\\ &= \sum_{n = 2}^∞ \frac{2^{3n}}{n^2} · 2^{-4n} \leqslant \sum_{n = 2}^∞ \frac{1}{n^2} < +∞. \tag*{$\square$} \end{align*}

Proposition 4: $g(x) + \dfrac{1}{p} f(x) = \dfrac{1}{p} h(x)$ for $x > 0$.

Proof: Note that\begin{gather*} g(x) = x^p \int_1^x \frac{f(u)}{u^{p + 1}} \,\d u = x^p \int_1^x \paren{ \frac{h(u)}{u^{p + 1}} - \frac{p}{u^{p + 1}} \int_1^u \frac{h(v)}{v} \,\d v } \,\d u. \tag{1} \end{gather*} Since\begin{gather*} \int_1^x \frac{p}{u^{p + 1}} \int_1^u \frac{h(v)}{v} \,\d v\d u = \int_1^x \frac{h(v)}{v} \,\d v \int_v^x \frac{p\,\d u}{u^{p + 1}} = \int_1^x \paren{ \frac{1}{v^p} - \frac{1}{x^p} } \frac{h(v)}{v} \,\d v, \end{gather*} plugging into (1) yields\begin{align*} g(x) &= x^p \paren{ \int_1^x \frac{h(u)}{u^{p + 1}} \,\d u - \int_1^x \frac{h(u)}{u^{p + 1}} \,\d u + \frac{1}{x^p} \int_1^x \frac{h(u)}{u} \,\d u }\\ &= \int_1^x \frac{h(u)}{u} \,\d u = \frac{1}{p} f(x) - \frac{1}{p} h(x). \tag*{$\square$} \end{align*}

Proposition 5: For any $ε > 0$, $h(x) ≠ O(x^ε)$ ($x → 0^+$).

Proof: Suppose $h(x) = O(x^ε)$ ($x → 0^+$) for some $ε > 0$, then there exist $δ > 0$ and $C > 0$ such that$$ h(x) = |h(x)| \leqslant C|x^ε|\ (x \in (0, δ)). $$ However, $h(2^{-n}) = n^{-2}$ for $n \geqslant 2$, a contradiction.

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