# Problem 11, Chapter 18 of Calculus by Spivak

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 ?

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.

• How are you going from $F(x) \geq \int_1^X \frac{f(t)}{t}dt + M \ln x - M \ln x$ to $\frac{F(x)}{\ln(x)} \geq \frac{C}{\ln x}+M$?
– S.C.
Apr 21, 2023 at 17:14
• given $M$, then $X$ is fixed, so the integral is a constant, and $-M\ln X$ is also a constant. Apr 22, 2023 at 11:05
• 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?
– S.C.
Apr 22, 2023 at 13:59
• 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? Apr 22, 2023 at 14:38
• 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 $f(1^+)$. Apr 22, 2023 at 14:38