Intereting Posts

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discrete version of Laplacian
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Let V denote the Klein 4-group. Show that $\text{Aut} (V)$ is isomorphic to $S_3$

I would appreciate if somebody could help me with the following problem:

Q: Let $f(x),f'(x),f”(x),f”'(x)>0$ , $f”'(x)$ is a continuous function and $f”'(x)<f(x)$ on $\mathbb{R}$

then show that

$$f'(x)<2f(x),~ \forall x\in \mathbb{R}$$

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- Present a function with specific feature
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From the assumptions we know that $f$, $f’$ and $f”$ are positive and increasing, so $$\lim_{x\to-\infty}f(x)\ge 0, \quad\text{and}\quad\lim_{x\to-\infty}f'(x)=\lim_{x\to-\infty} f”(x)=0.$$

It follows that

$$f”(x)=\int_{-\infty}^xf”'(t) \, dt,$$

$$f'(x)=\int_{-\infty}^xf”(t) \, dt=\int_{-\infty}^x \left(\int_{-\infty}^tf”'(s) \, ds \right) \, dt=\int_{-\infty}^x(x-s)f”'(s) \, ds,\tag{1}$$

and

$$f(x)\ge\int_{-\infty}^xf'(t) \, dt=\int_{-\infty}^x \left(\int_{-\infty}^t (t-s) f”'(s) \, ds \right) \, dt=\frac{1}{2}\int_{-\infty}^x(x-s)^2f”'(s)\,ds.\tag{2}$$

Fix some $\lambda>0$ to be determined below. Then, since $x-s<\frac{(x-s)^2}{\lambda}$ for $s\in(-\infty,x-\lambda)$,

$$\begin{eqnarray*}

f'(x) &\le &\lambda^{-1}\int_{-\infty}^{x-\lambda}(x-s)^2f”'(s)ds+\int_{x-\lambda}^x(x-s)f”'(s)ds \tag{by (1)}\\

&\le & 2\lambda^{-1}f(x)+\int_{x-\lambda}^x [(x-s)-\lambda^{-1}(x-s)^2]f”'(s) \, ds\tag{by (2)}\\

&<& 2\lambda^{-1}f(x)+f(x)\int_{x-\lambda}^x[(x-s)-\lambda^{-1}(x-s)^2] \, ds\tag{$f”'(s)<f(x)$}\\

&=&\left(2\lambda^{-1}+\frac{\lambda^2}{6}\right)f(x).

\end{eqnarray*}$$

Since $\min_{\lambda>0}(2\lambda^{-1}+\frac{\lambda^2}{6})=\frac{3}{\sqrt[3]{6}}<2$( for $\lambda=\sqrt[3]{6}$), the conclusion follows.

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