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The problem:
Let $f$ be a nondecreasing function on $[1;\infty)$. Define $F(x)$ as follows: $$F(x) = \int_{1}^{x}\frac{f(t)}{t}dt.$$ Prove that $f$ is bounded on $[1;\infty)$ if only if $\frac{F}{\log}$ is bounded on $[1;\infty)$.

The left-to-right can be easily proved. Let's just consider the converse.

Here's what i did:
Since $f$ is nondecreasing, $\forall x\in[1;\infty]$: We have: $\forall t\in[1;x]: f(t) \leq f(x)$.
So, $$F(x)=\int_{1}^{x}\frac{f(t)}{t}dt \leq \int_{1}^{x}\frac{f(x)}{t}dt = f(x).log(x) (*)$$ (In Spivak's Calculus, he defines $log(x) = \int_{1}^{x}\frac{1}{t}dt$)

If $f<0$ on $[1;+\infty)$ then: $$F(x) = \int_{1}^{x}\frac{f(t)}{t}dt < \int_{1}^{x}0dt = 0$$. And: $(*)$ becomes: $$-|F(x)| \leq -|f(x)|.logx \iff |F(x)| \geq f(x).logx \iff |\frac{F(x)}{logx}|\geq|f(x)|$$. Since $F/log$ bounded, so is $f$
If $\exists x\in[1;+\infty): f(x) \geq 0$ then the set $A=\{x\geq 1:f<0\ \text{on}\ [1;x]\}$ must be bounded above.

Consider $\alpha = \text{sup}A$.
So, $f<0$ on $[1;\alpha]$ but $\geq 0$ on $(\alpha;+\infty)$. Similarly, we can prove $f$ is bounded on $[1;\alpha]$. But i'm not quite sure how to handle the $(\alpha;+\infty)$. On $(\alpha;+\infty)$, what we have is: $$F(x)=(\int_{1}^{\alpha}+\int_{\alpha}^{x})\frac{f(t)}{t}dt \leq c_0 + f(x)(logx-log(\alpha)) \iff |\frac{F(x)}{logx - log\alpha}| - \frac{c_0}{logx - log\alpha} \leq |f(x)|$$. We cannot conclude $f$ is bounded at all. And the relation "$\leq$" seems to be the only way to take out $f(x)$. Any ideas on how to proceed ?

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1 Answer 1

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Why not prove it by contradiction?

If $f$ is not bounded, then for any large $M > 0$ there exists $X$ such that $f(x) \geq M$ for all $x\geq X$. This means that $F(x)\geq \int_1^X \frac{f(t)}{t} dt + M\ln x - M\ln X$. Thus $$ \frac{F(x)}{\ln x} \geq \frac{C}{\ln x} + M $$ where $C$ is a constant depending on $M$ and $X$. Letting $x\to\infty$, we see that $F(x)/\ln x \geq M/2$ for all large $x$. This result holds for any $M$, and thus $F(x)/\ln x$ cannot be bounded.

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  • $\begingroup$ How are you going from $F(x) \geq \int_1^X \frac{f(t)}{t}dt + M \ln x - M \ln x$ to $\frac{F(x)}{\ln(x)} \geq \frac{C}{\ln x}+M$? $\endgroup$
    – S.C.
    Apr 21, 2023 at 17:14
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    $\begingroup$ given $M$, then $X$ is fixed, so the integral is a constant, and $-M\ln X$ is also a constant. $\endgroup$
    – user58955
    Apr 22, 2023 at 11:05
  • $\begingroup$ Thank you for the response. One thing is confusing me, though. In the assumption '$f$ is not bounded', you implicitly assumed that $f$ is not bounded from above. However, what if $f$ is bounded from above, but not bounded from below. At first I dismissed this case because the problem used the closed bracket around $1$ in $[1,\infty)$. However, $\frac{1}{\log x}$ is not defined at $x=1$, which leads me to believe that it is perfectly reasonable to have also assumed that $f$ is not defined at $x=1$ either. Shouldn't this circumstance also be considered? $\endgroup$
    – S.C.
    Apr 22, 2023 at 13:59
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    $\begingroup$ Oh, that's a good point. Note that $F(x)/\ln x = \frac{1}{\ln x}\int_1^x \frac{f(t)}{t}dt \geq \frac{1}{\ln x}\int_1^x \frac{f(1^+)}{t}dt = f(1^+)$, it is always bounded from below. For $x=1$, you can take lim inf and the lim inf is at least $f(1^+)$. On the other hand, when $x<x_0$, $F(x)/\ln x \leq \frac{1}{\ln x}\int_1^x \frac{f(x_0)}{t}dt \leq f(x_0)$, so lim sup is at most $f(x_0)$ for any $x_0 > 1$, that is, the lim sup is at most $f(1^+)$. Does this clarify your question? $\endgroup$
    – user58955
    Apr 22, 2023 at 14:38
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    $\begingroup$ As to your other question, it's the wording that $f$ is non-decreasing on $[1,\infty)$ which demends that $f$ is defined at $1$. But here it doesn't really matter. You can replace $f(1)$ with $f(1^+)$. $\endgroup$
    – user58955
    Apr 22, 2023 at 14:38

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