Does this condition on the sum of a function and its integral imply that the function goes to 0?

Question

Consider a bounded real-valued function $S:\mathbf{R}\to\mathbf{R}$ so that

$$\lim_{x\to\infty} \left( S(x) + \int_1^x \frac{S(t)}{t}dt\right)$$

exists and is finite. Can one say that $\lim_{x\to\infty} S(x)=0$?

Answer 1

Following from Greg Martin's comment :

$\lim\limits_{x\to \infty} T(x)+\int\limits_{0}^xT(u)\,du = l$ (exists !)

Assuming $T$ is continuous,

Write it as, $F(x)=\int\limits_{0}^xT(u)\,du$,and we get $\lim\limits_{x\to \infty} F(x)+T(x)=l$.

Given an $\epsilon>0,\exists a>0,$ such that $x>a \implies |F(x)+T(x)-l|<\epsilon$.

From generalized Mean Value Theorem, $\exists \omega \in (a,x)$ such that, $\dfrac{e^xF(x)-e^aF(a)}{e^x-e^a}=F(\omega)+T(\omega)$.

or, $(l-\epsilon)<\dfrac{e^xF(x)-e^aF(a)}{e^x-e^a}<(l+\epsilon)$

or, $(l-\epsilon)(1-e^{(a-x)})<F(x)-e^{(a-x)}F(a)<(1-e^{(a-x)})(l+\epsilon)$

or, $e^{(a-x)}|F(a)|+(l-\epsilon)(1-e^{(a-x)})<F(x)< e^{(a-x)}|F(a)|+(l+\epsilon)(1-e^{(a-x)})$

Therefore, $\lim\limits_{x\to \infty}F(x)=l$.

i.e., $\lim\limits_{x\to \infty}T(x)=\lim\limits_{x\to \infty}T(x)+F(x)-F(x)=l-l=0$.

Answer 2

I will assume that $S \in L^{\infty}(\Bbb{R})$. Let $T(x)$ by

$$ T(x) = x \int_{1}^{x} \frac{S(t)}{t} \, dt. $$

Then by the Lebesgue differentiation theorem, $T(x)$ is a.e. differentiable and

$$ T'(x) = S(x) + \int_{1}^{x} \frac{S(t)}{t} \, dt. $$

Now by the assumption, $T'(x) \to \ell$ for some real number $\ell$ as $x \to \infty$. Then it is easy to prove that

$$ \int_{1}^{x} \frac{S(t)}{t} \, dt = \frac{T(x)}{x} \to \ell $$

as well! Therefore

$$ S(x) = T'(x) - \frac{T(x)}{x} \to \ell - \ell = 0 $$

as $x \to \infty$.

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Addendum. Let us prove the following lemma.

Lemma. $T'(x) \to \ell$ as $x \to \infty$, then $T(x)/x \to \ell$ as $x \to \infty$.

Indeed, we can use the following strategy which is standard in Abelian theorems. (As you see, this lemma says that ordinary limit implies Cesàro mean.)

Proof. For any $\epsilon > 0$, choose $R > 0$ such that $|T'(x) - \ell| < \epsilon$ whenever $x \geq R$. Then

\begin{align*} \frac{T(x)}{x} - \ell &= \frac{1}{x} \left( T(R) + \int_{R}^{x} T'(t) \, dt \right) - \ell \\ &= \frac{1}{x} \left( T(R) - R\ell + \int_{R}^{x} (T'(t) - \ell) \, dt \right). \end{align*}

Thus taking absolute value,

\begin{align*} \left| \frac{T(x)}{x} - \ell \right| &\leq \frac{\left| T(R) - R\ell \right|}{x} + \frac{1}{x} \int_{R}^{x} \left| T'(t) - \ell \right| \, dt \\ &\leq \frac{\left| T(R) - R\ell \right|}{x} + \frac{x - R}{x} \, \epsilon. \end{align*}

Taking $x \to \infty$, we have

$$ \limsup_{x\to\infty} \left| \frac{T(x)}{x} - \ell \right| \leq \epsilon. $$

But since this is true for any $\epsilon > 0$, the left-hand side vanishes and the proof is complete. ////