17
$\begingroup$

Question

Let $\lambda \in (0,1), s \in (0,\infty), d \in \{2,3,\dots\}$ and show that in this case the following inequality holds: $$(\lambda^d + (1-\lambda^d) e^{(d-1)s})^{\frac{1}{1-d}} \leq \sum_{n=0}^\infty \frac{\lambda^{\frac{(d^n-1)d}{d-1}+n}s^n}{n!} e^{-\lambda s},$$

Thoughts

My first intuition was to try and write the left hand side as some taylor series but this seems like a bad idea as the only part to write as a Taylor series is $e^{(d-1)s}$ and if we "ignore" the $\lambda^d$ this part is what gives $e^{-s}$ which is bounded by $e^{-\lambda s}$.

Maybe there is some formula for a convex expression of the form $\lambda^d a + (1-\lambda^d) b$ which can be used here? Applying convexity I can write the right hand side as $e^{-(1-\lambda^d)s}$ which, unfortunately, is not smaller than the right hand side (for large values of $s$ the inequality fails, which is logical as convecity inequality is increasingly inaccurate as $s$ increases).

I might be able to show it using the Taylor expansion of $f(x) := \frac{1}{(1+x)^{1/n}}$ as I can write the right hand side as: $$ (1 + (1-\lambda^d) (e^{(d-1)s}-1))^{1/(1-d)}, $$ for this I can take the taylor series at $0$ for $|x|=|(1-\lambda^d) (e^{(d-1)s}-1)| < 1$ and at $\infty$ for $|x| > 1$.

It seems to me that I need something like the binomial approximation but for large values of $x$ rather than small values of $x$.

Promising technique

I can write the left hand side as $$e^{-s} (1 + \lambda^d (e^{(1-d)s} - 1))^{1/(1-d)}$$ as $|\lambda^d (1-e^{(1-d)s})| < 1$ we can use the Taylor expansion of $1/(1+x)$ at $0$. This allows us to reduce the inequality to showing: $$ e^{-s} \sum_{n=0}^{\infty} \frac{1}{n!} \frac{\Gamma(1 - 1/(d-1))}{\Gamma(1 - 1/(d-1) - n)} \cdot \lambda^{nd} (e^{(1-d)s}-1)^n \leq \sum_{n=0}^{\infty} \frac{\lambda^{((d^n-1)d)/(d-1)+n}s^n}{n!}e^{-\lambda s}. $$ The separate terms in the sum on the right hand side drops to zero much faster than the sum on the left hand side, but the first couple of terms of the right hand side are a lot larger than those on the left hand side. It seems to me that this corresponds to the fact that the Taylor series on the left hand side converges very slowly to the function, maybe we can take a better Taylor series which converges faster.

$\endgroup$
  • $\begingroup$ Hello , maybe you could use the $(\alpha,m)-convex$ definition 1.2 of this link . With this definition we have : $$ f(t^d+(1-t^d)y)\leq tf(x)+(1-t)f(y)$$ where $d=\frac{1}{\alpha}$ and $m=1$ . Furthermore you have in this link some results related to $(\alpha,m)-convex$ .Good luck . $\endgroup$ – max8128 Jan 23 '18 at 10:50
  • $\begingroup$ I will look into this! Thank you for the response. $\endgroup$ – HolyMonk Jan 23 '18 at 10:51
  • $\begingroup$ @max8128 : This technique (although interesting) does not seem that helpful here, the notion of $(1,1/d)$ convexity is equivalent to $f'(x) (x-y) \geq \frac{1}{d} (f(x) - f(y))$ and this is weaker than normal convexity, therefore the properties we would employ for $(1,1/d)$ convexity would be better to just apply the analogues for normal convexity as it is stronger. The problem with just using $f(t^dx+(1-t^d)y) \leq t^d f(x) + (1-t^d) f(y)$ is not the $d-$power of $t$, but rather the fact that in our case we use this with $x = 0$ and $y = (d-1)s$, thus $x = 0$ and $y$ is very big, making this $\endgroup$ – HolyMonk Jan 23 '18 at 11:09
  • $\begingroup$ @max8128 inequality a very crude estimate. $\endgroup$ – HolyMonk Jan 23 '18 at 11:09
14
+100
$\begingroup$

$\def\e{\mathrm{e}} \def\veq{\mathrel{\phantom{=}}}$Denote $μ = λ^d$. It will be proved that\begin{align*} (μ + (1 - μ) \e^{(d - 1)s})^{-\frac{1}{d - 1}} \leqslant \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}} \end{align*} holds for $0 < μ < 1$, $s > 0$ and $d > 1$. Step 1 and Step 2 prepare for the proof of monotonicity in Step 3.

Lemma: For $0 < x <1$ and $r > 1$,$$ 1 - x^r \leqslant r(1 - x). $$ (This is trivial by considering the partial derivative of $x^r - rx$ with respect to $x$.)

Step 1: For $0 < μ < 1$, $d > 1$ and $n \in \mathbb{N}$,\begin{align*} &\veq \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k}))\\ &\geqslant μ \cdot (μ^{\frac{1}{d}} + d - 1)^n \cdot μ^{\frac{d^n - 1}{d - 1}} \geqslant μ \cdot μ^{\frac{n}{d}} d^n μ^{\frac{d^n - 1}{d - 1}} \geqslant μ \cdot μ^{\frac{n}{d}} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}}. \tag{1.1} \end{align*}

Proof: For $n = 0$, (1.1) becomes$$ 1 - μ^{\frac{1}{d}} (1 - μ) \geqslant μ \geqslant μ, $$ which is true because$$ 1 - μ^{\frac{1}{d}} (1 - μ) - μ = (1 - μ)(1 - μ^{\frac{1}{d}}) \geqslant 0. $$

For $n \geqslant 2$, since$$ (μ^{\frac{1}{d}} + d - 1)^n = \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k}, $$ then\begin{align*} &\veq \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ \cdot (μ^{\frac{1}{d}} + d - 1)^n μ^{\frac{d^n - 1}{d - 1}}\\ &= \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} \left( μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right). \tag{1.2} \end{align*}

For the $k = n$ term in (1.2), by the lemma,\begin{align*} &\veq μ^{\frac{n}{d}} \left( μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^n})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right) = μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ - μ^{\frac{1}{d}} (1 - μ^{d^n}))\\ &\geqslant μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ - μ^{\frac{1}{d}} d^n (1 - μ)) = μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ)(1 - μ^{\frac{1}{d}} d^n)\\ &\geqslant μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ)(1 - d^n) = - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ) \sum_{k = 0}^{n - 1} \binom{n}{k} (d - 1)^{n - k}, \end{align*} thus\begin{align*} (1.2) &\geqslant \sum_{k = 0}^{n - 1} \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} \left( μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right)\\ &\veq - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ) \sum_{k = 0}^{n - 1} \binom{n}{k} (d - 1)^{n - k}\\ &= \sum_{k = 0}^{n - 1} \binom{n}{k} (d - 1)^{n - k} \left( μ^{\frac{k}{d}} \left( μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right) - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ) \right). \tag{1.3} \end{align*}

For $0 \leqslant k \leqslant n - 1$, since $0 < μ < 1$ and $d > 1$, then $μ^{\frac{k}{d}}$, $μ^{\frac{d^k - 1}{d - 1}}$ and $μ^{d^k}$ are all decreasing with respect to $k$. Thus\begin{align*} &\veq μ^{\frac{k}{d}} \left( μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right) - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ)\\ &\geqslant μ^{\frac{n - 1}{d}} \left( μ^{\frac{d^{n - 1} - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^{n - 1}})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right) - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ)\\ &= μ^{\frac{n - 1}{d}} μ^{\frac{d^{n - 1} - 1}{d - 1}} \left( (1 - μ^{\frac{1}{d}} (1 - μ^{d^{n - 1}})) - μ^{d^{n - 1}} \cdot μ - μ^{\frac{1}{d}} μ^{d^{n - 1}} (1 - μ) \right)\\ &= μ^{\frac{n - 1}{d}} μ^{\frac{d^{n - 1} - 1}{d - 1}} \left( 1 - μ^{\frac{1}{d}} - μ^{d^{n - 1} + 1} + μ^{\frac{1}{d} + d^{n - 1} + 1}\right)\\ &= μ^{\frac{n - 1}{d}} μ^{\frac{d^{n - 1} - 1}{d - 1}} (1 - μ^{\frac{1}{d}})(1 - μ^{d^{n - 1} + 1}) \geqslant 0, \end{align*} which implies $(1.3) \geqslant 0$. Therefore,$$ \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) \geqslant μ \cdot (μ^{\frac{1}{d}} + d - 1)^n μ^{\frac{d^n - 1}{d - 1}}. $$ Also, $0 < μ < 1$ and $d > 1$ implies $μ^{\frac{1}{d}} + d - 1 \geqslant d μ^{\frac{1}{d}}$, and by the lemma, $\displaystyle d^n \geqslant \frac{1 - μ^{d^n}}{1 - μ}$. Hence (1.1) holds for $0 < μ < 1$, $d > 1$ and $n \in \mathbb{N}$.

Step 2: For $0 < μ < 1$, $s > 0$ and $d > 1$,\begin{align*} &\veq \e^{(d - 1)s} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^n}))\\ &\geqslant μ \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}} \geqslant μ^{\frac{1}{d}} μ \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}}. \tag{2.1} \end{align*}

Proof: By Mertens' theorem,\begin{align*} &\veq \e^{(d - 1)s} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^n}))\\ &= \left( \sum_{n = 0}^\infty \frac{((d - 1)s)^n}{n!} \right) \left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^n})) \right)\\ &= \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{((d - 1)s)^{n - k}}{(n - k)!} \cdot \frac{(sμ^{\frac{1}{d}})^k}{k!} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^k}))\\ &= \sum_{n = 0}^\infty \frac{s^n}{n!} \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^k})). \tag{2.2} \end{align*}

By (1.1),\begin{align*} (2.2) &\geqslant \sum_{n = 0}^\infty \frac{s^n}{n!} μ \cdot μ^{\frac{n}{d}} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}} = μ \cdot \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}}\\ &\geqslant μ^{\frac{1}{d}} μ \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}}. \end{align*} Hence (2.2) holds for $0 < μ < 1$, $s > 0$ and $d > 1$.

Step 3: For $0 < μ < 1$ and $d > 1$,\begin{align*} f(s) &= \ln\left( \left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}} \right) (μ + (1 - μ) \e^{(d - 1)s})^{\frac{1}{d - 1}} \right)\\ &= \ln\left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \right) - sμ^{\frac{1}{d}} + \frac{1}{d - 1} \ln(μ + (1 - μ) \e^{(d - 1)s}) \end{align*} is increasing for $s > 0$.

Proof: Because for any $A > 0$, the series$$ \sum_{n = 0}^\infty \left( \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \right)' = \sum_{n = 1}^\infty \frac{n s^{n- 1} μ^{\frac{n}{d}}}{n!} μ^{\frac{d^n - 1}{d - 1}} = μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^{n + 1} - 1}{d - 1}} $$ converges uniformly for $s \in (0, A)$, then for any $s > 0$,\begin{align*} f'(s) &= \frac{\sum\limits_{n = 0}^\infty \left( \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \right)'}{\sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}}} - μ^{\frac{1}{d}} + \frac{1}{d - 1} \frac{(μ + (1 - μ) \e^{(d - 1)s})'}{μ + (1 - μ) \e^{(d - 1)s}}\\ &= \frac{μ^{\frac{1}{d}} \sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^{n + 1} - 1}{d - 1}}}{\sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}}} - μ^{\frac{1}{d}} + \frac{(1 - μ) \e^{(d - 1)s}}{μ + (1 - μ) \e^{(d - 1)s}}. \end{align*}

Define$$ A = μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} - μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^{n + 1} - 1}{d - 1}}, \quad B = \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}}. $$ Because\begin{align*} A &= μ^{\frac{1}{d}} \left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} - \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^{n + 1} - 1}{d - 1}} \right)\\ &= μ^{\frac{1}{d}} \left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} - \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \cdot μ^{d^n} \right)\\ &= μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{d^n}), \end{align*}\begin{align*} B - A &= \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} - μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{d^n})\\ &= \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^n})), \end{align*} then\begin{align*} f'(s) \geqslant 0 &\Longleftrightarrow \frac{(1 - μ) \e^{(d - 1)s}}{μ + (1 - μ) \e^{(d - 1)s}} \geqslant \frac{A}{B}\\ &\Longleftrightarrow B(1 - μ) \e^{(d - 1)s} \geqslant A(μ + (1 - μ) \e^{(d - 1)s})\\ &\Longleftrightarrow (1 - μ)(B - A) \e^{(d - 1)s} \geqslant μA\\ &\Longleftrightarrow (B - A) \e^{(d - 1)s} \geqslant \frac{μ}{1 - μ} A\\ &\Longleftrightarrow \e^{(d - 1)s} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^n})) \geqslant μ^{\frac{1}{d}} μ \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \frac{1 - μ^{d^n}}{1 - μ}, \end{align*} where the last inequality holds by (2.1). Hence $f(s)$ is increasing for $s > 0$.

Step 4: For $0 < μ < 1$, $s > 0$ and $d > 1$,\begin{align*} (μ + (1 - μ) \e^{(d - 1)s})^{-\frac{1}{d - 1}} \leqslant \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}}. \tag{4.1} \end{align*}

Proof: From Step 3 it is known that $f(s)$ is increasing for $s > 0$, thus$$ \exp(f(s)) = \frac{\sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}}}{(μ + (1 - μ) \e^{(d - 1)s})^{-\frac{1}{d - 1}}} $$ is also increasing for $s > 0$. Because the series $\sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}}$ converges uniformly for $s \in [0, 1]$, then for any $s_0 > 0$,\begin{align*} &\veq \frac{\sum\limits_{n = 0}^\infty \frac{(s_0 μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s_0 μ^{\frac{1}{d}}}}{(μ + (1 - μ) \e^{(d - 1)s_0})^{-\frac{1}{d - 1}}} = \exp(f(s_0)) \geqslant \lim_{s \to 0^+} \exp(f(s))\\ &= \frac{\lim\limits_{s \to 0^+} \sum\limits_{n = 0}^\infty \frac{(s μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s μ^{\frac{1}{d}}}}{\lim\limits_{s \to 0^+} (μ + (1 - μ) \e^{(d - 1)s})^{-\frac{1}{d - 1}}} = \frac{\sum\limits_{n = 0}^\infty \lim\limits_{s \to 0^+} \frac{(s μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s μ^{\frac{1}{d}}}}{\left( μ + (1 - μ) \lim\limits_{s \to 0^+} \e^{(d - 1)s} \right)^{-\frac{1}{d - 1}}}\\ &= \frac{\lim\limits_{s \to 0^+} \e^{-s μ^{\frac{1}{d}}} + \sum\limits_{n = 1}^\infty \lim\limits_{s \to 0^+} \frac{(s μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s μ^{\frac{1}{d}}}}{\left( μ + (1 - μ) \lim\limits_{s \to 0^+} \e^{(d - 1)s} \right)^{-\frac{1}{d - 1}}} = \frac{1}{(μ + (1 - μ))^{-\frac{1}{d - 1}}} = 1, \end{align*} i.e.$$ \sum_{n = 0}^\infty \frac{(s_0 μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s_0 μ^{\frac{1}{d}}} \geqslant (μ + (1 - μ) \e^{(d - 1)s_0})^{-\frac{1}{d - 1}}. $$ Hence (4.1) holds for $0 < μ < 1$, $s > 0$ and $d > 1$.

$\endgroup$
  • $\begingroup$ Wow, thank you! I will read your answer carefully now, looks very impressive! $\endgroup$ – HolyMonk Jan 27 '18 at 10:48
  • $\begingroup$ I went through it and I approve! Thanks again great answer! $\endgroup$ – HolyMonk Jan 27 '18 at 14:04
  • $\begingroup$ Wow, +1 is not enough to express the degree of impressiveness of this proof! $\endgroup$ – Hans Jan 30 '18 at 20:05
2
$\begingroup$

Too long for a comment (but I can delete it if you want) so :

In fact the RHS is a polynomial more particulary a Jensen polynomial :

Why? Because the coefficients fulfill the conditions also called Turan's inequality: $$\sigma_k^2\geq \sigma_{k-1}\sigma_{k+1}$$

Proof (here $\lambda=\sigma$): We have to prove :

$$\sigma^{2\frac{d^n-1}{d-1}+2n}\geq \sigma^{\frac{d^{n-1}-1}{d-1}+n-1}\sigma^{\frac{d^{n+1}-1}{d-1}+n+1}$$

After simplification we have :

$$\sigma^{2\frac{d^n-1}{d-1}}\geq\sigma^{\frac{d^{n-1}-1}{d-1}}\sigma^{\frac{d^{n+1}-1}{d-1}}$$ We take the logarithm on each side we get : $${2\frac{d^n-1}{d-1}}ln(\sigma)\geq\frac{d^{n-1}-1}{d-1}ln(\sigma)+\frac{d^{n+1}-1}{d-1}ln(\sigma)$$

The logaritm is negative so we reverse the inequality :

$${2(\frac{d^n-1}{d-1})}\leq\frac{d^{n-1}-1}{d-1}+\frac{d^{n+1}-1}{d-1}$$

Finally we get :

$$2\leq \frac{1}{d}+d$$ Wich is obvious

So it's a serious way to prove it but now I have not the time to conclude

Edit : The associated polynomials to Jensen polynomials are :

$$g_n(t)=\sum_{k=0}^{n}{n\choose k}\sigma_tx^t$$

Futhermore we know (let $g_n(t)=g_n(t)(\phi;x)$) then the polynomials $g_n(t)$ are generated by : $$e^t\phi(xt)=\sum_{k=0}^{\infty}g_n(x)\frac{t^n}{n!}$$ Theorem A from this link asserts this : the function $\phi(x)$ belongs to L-P if and only if all the polynomials $g_n(\phi;x)$,$n=1,2,\cdots$ have only real zeros .

From this link we can say that the zeros of Jensen polynomials are simple and negative so we can express the Jensen polynomials like this :

$$f(z)=cz^me^{-\alpha z^2+\beta z }\prod_{n=0}^{\infty}(1-\frac{z}{z_n})e^{\frac{z}{z_n}}$$

So the infinite sum becomes a product... maybe it could help .

$\endgroup$
  • $\begingroup$ I am definitely intrigues by this approach, I am now googling for special properties of Jensen polynomials! $\endgroup$ – HolyMonk Jan 24 '18 at 9:38
  • $\begingroup$ Small comment: The right hand side should (I think) be written as: $\left(\sum_{n=0}^\infty \lambda ^{\frac{(d^n-1)d}{d-1} + n} \frac{s^n}{n!}\right) e^{-\lambda s}$ and we then take $\gamma_n = \lambda ^{\frac{(d^n-1)d}{d-1} + n}$. (just to be $100\%$ clear) $\endgroup$ – HolyMonk Jan 24 '18 at 9:56
  • $\begingroup$ I did not succeed at continuing from this. $\endgroup$ – HolyMonk Jan 25 '18 at 8:25
  • $\begingroup$ @Holy Monk See my edits . Your inequality I think is very hard ^^ .Have a good day . $\endgroup$ – max8128 Jan 25 '18 at 10:45
  • $\begingroup$ I think so too, thanks for trying I will be looking more into your approach as it does interest me, if I find it using your approach I will let you know and accept your answer. $\endgroup$ – HolyMonk Jan 25 '18 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.