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I am currently studying Mathematical Analysis and I wonder if such a proposition holds :

If $\ a,c>0 , b>1$, and $a<c \ b^{1/n}$, then $a \le c\ \inf_{n \in N}b^{1/n}$

Originally my question is from the proposition that

If $\ b^{x+y} < b^x b^yb^{1/n}, \ $ then $\ b^{x+y} \le b^x b^y\ \inf_{n \in N}b^{1/n} $.

How does it work? I know it's naively true if we consider the situation $n \rightarrow \infty$, but I want rigorous proof.

이승우
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2 Answers2

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Yes it is true. First note that $b^\frac{1}{n} \in (\frac{a}{c}, \infty)$ for all $n \in \mathbb{N}$ and thus the infimum is finite. So, by definition of the infimum, there exists a sequence $(n_k)_{k \in \mathbb{N}}\subseteq \mathbb{N}$ such that $$ \lim_{k \rightarrow \infty} b^\frac{1}{n_k} = \inf_{n \in \mathbb{N}} b^\frac{1}{n}. $$ As $\inf$ is the limit of a sequence in $(\frac{a}{c}, \infty)$, it will live in the closure, i.e. $[\frac{a}{c}, \infty)$. As it is finite, we have $$ \inf_{n \in \mathbb{N}} b^\frac{1}{n} \geq \frac{a}{c} \iff c \inf_{n \in \mathbb{N}} b^\frac{1}{n} \geq a. $$

Hyperbolic PDE friend
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I'm not sure if it is just a typo, but if you have $a/c < b^{1/n}$ for all $n$, then $a\leq c$, since $\inf_n b^{1/n} = 1$. This is because $b^{1/n}$ is decreasing with respect to $n$ since $b>1$ (see it as $b^{1/n} = e^{\log(b) / n}$), and $\lim_{n \to \infty} b^{1/n} = 1$.

SoufHay
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  • If we don't have $\inf_n b^{1/n} =1 $, do we still get $a \le c\ \inf_{n \in N}b^{1/n}$ ? – 이승우 Apr 10 '21 at 12:55
  • Yes, the statement $a \leq c \inf_n b^{1/n}$ is always true wether $b>1$ or not. However, $b^{1/n}$ is only defined when $b>0$. Thus, either $b > 1$ in which case $\inf_n b^{1/n} = 1$ or $b \in (0,1)$ in which case $\inf_{n \in \mathbb{N}^*} b^{1/n} = b$ (taking $n=1$). – SoufHay Apr 10 '21 at 13:00
  • But, why is $a \leq c \inf_n b^{1/n}$ always true? I think my original question can be revised as follows : If we have $a_n < b_n$ for all $n \in N$, do we always have $\inf {a_n} \le \inf {b_n}$? If it is, how can we prove it ?? My point is that how < can be expanded to $\le$. – 이승우 Apr 10 '21 at 13:10
  • It's a simple result. If $a_n<b_n$, then $\inf_k a_k < b_n$ since by definition $a_n \geq \inf_k a_k$. Now $\inf_k a_k < b_n$ is true for all $n$, which yields $\inf_k a_k \leq \inf_n b_n$. The last point is trivial, but if you want to prove it, consider a sequence $\alpha(n)$ such that $\lim_n b_{\alpha(n)} = \inf_k b_k$ ($\alpha$ exists by definition of inf), and take the limit. – SoufHay Apr 10 '21 at 13:28
  • Ahh okay. it is really simple. Thank you so much for detailed answer. – 이승우 Apr 10 '21 at 13:39