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I was reading this question Showing this to $1 \leq p < \infty$ , here on this site.

But I did not understand this part (in the third solution mentioned there given by @mechanodroid):

" Now notice that for large enough $t_0 \ge 1$ we have $t \ge t_0 \implies \ln t \le \sqrt[2p]{t}$ so the above integral can be bounded by $$\int_1^{t_0}\frac{(\ln t)^p}{t^2}\,dt + \int_{t_0}^{+\infty}\frac{dt}{t^{3/2}} < +\infty$$ which is finite since $\frac{(\ln t)^p}{t^2}$ is bounded on $[1,t_0]$ and $t \mapsto \frac1{t^{3/2}}$ is integrable on $[1, +\infty)$."

Could anyone clarify this part for me, please?

2- Also for the second solution given by @Awegan, I can see a typo in $p$ and $q$, the powers of $L$, am I correct?

3- Also, is there an alternative proof (detailed) for the question given in the link I mentioned above?

1 Answers1

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$$\int_1^{t_0}\frac{(\ln t)^p}{t^2}\,dt + \int_{t_0}^{+\infty}\frac{dt}{t^{3/2}} < +\infty$$

Since the integrand of the left hand integral is bounded, then $$\int_1^{t_0}\frac{(\ln t)^p}{t^2}dt\leq M(t_0-1),$$ where $M$ is the bound (see here). Clearly this is finite.

For the right hand integral, just evaluate it, e.g.

$$\int_{t_0}^\infty\frac{dt}{t^{3/2}}dt=\left[-\frac{2}{\sqrt{t}}\right]_{t_0}^{+\infty}=\frac{2}{\sqrt{t_0}},$$ which is also finite.

The sum of two finite numbers gives a finite number, so by definition of $L^p$ spaces $f\in L^p(0,1]$.

In answer to your comment, here's a plot of $\log t$ (in blue) and $t^{1/(2p)}$ for $p=1.4$, for example. As you can see, $t_0\approx 35$ in this case, after which the inequality holds true for $t\geq t_0$. This always happens for some $t_0$ value.

enter image description here

pshmath0
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  • What about the rest of my questions? –  Nov 26 '19 at 10:35
  • @Mathstupid re (2) I think you may be right, there is a typo, check out this for comparison: https://math.stackexchange.com/questions/390227/proving-that-lp-subset-lq-when-1-le-q-le-p – pshmath0 Nov 26 '19 at 10:48
  • And why we are assuming "for large enough $t_0 \ge 1$" ? –  Nov 26 '19 at 10:51
  • Why "$\ln t \le \sqrt[2p]{t}$"? –  Nov 26 '19 at 11:00
  • Is there any mathematical proof for my last question other than the drawing? –  Nov 26 '19 at 11:11
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    @Mathstupid yes, check out https://math.stackexchange.com/questions/1663818/does-the-logarithm-function-grow-slower-than-any-polynomial – pshmath0 Nov 26 '19 at 11:47
  • why the integrand of the left hand integral is bounded? –  Nov 26 '19 at 15:25
  • Is this because it is a continuous function (as it is a rational function of a continuous ln function in the numerator and a polynomial (continuous) in the denominator) on a closed interval ? –  Nov 26 '19 at 15:28
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    @Mathstupid pretty much, yes. I mean $t>1$ so $\ln t$ behaves nicely, so that you get a continuous function $(\ln t)^p/t^2$, so since we consider integrating over $(1,t_0)$ then there must be some maximum $M$ of that function in $(1,t_0)$. – pshmath0 Nov 26 '19 at 16:23