Post History
#2: Post edited
by
Peter Taylor
·
2022-05-19T15:41:49Z (over 2 years ago)
Improve legibility
Consider $g^{-\epsilon}$. Then [its first derivative](https://youtu.be/mV0N3qHc4Bo?t=318) is $\color{limegreen}{D_x \\; g^{-\epsilon} = \epsilon g^{-1-\epsilon} g'}$. Then $g^{-\epsilon} > 0$ and tends to 0, and ${\color{limegreen}{D_x \\; g^{-\epsilon}}} < 0 $.If ${\color{limegreen}{D_x \\; g^{-\epsilon}}} < -\epsilon$ on a set of infinite measure, then $\int^{x}_0 {\color{limegreen}{D_x \, g^{-\epsilon}}} \quad dx $ (which differs from $g^{-\epsilon}$ by a constant) would diverge to negative infinity. So ${\color{limegreen}{D_x \\; g^{-\epsilon}}} \ge -\epsilon$ outside a set of finite measure, from which your desired inequality follows immediately.
- Consider $g^{-\epsilon}$. Then [its first derivative](https://youtu.be/mV0N3qHc4Bo?t=318) is $D_x \\; g^{-\epsilon} = \epsilon g^{-1-\epsilon} g'$. Then $g^{-\epsilon} > 0$ and tends to 0, and ${D_x \\; g^{-\epsilon}} < 0 $.
- If $D_x \\; g^{-\epsilon} < -\epsilon$ on a set of infinite measure, then $\int^{x}_0 D_x \, g^{-\epsilon} \quad dx $ (which differs from $g^{-\epsilon}$ by a constant) would diverge to negative infinity. So $D_x \\; g^{-\epsilon} \ge -\epsilon$ outside a set of finite measure, from which your desired inequality follows immediately.
#1: Initial revision
by
DNB
·
2022-05-15T19:15:28Z (over 2 years ago)
Consider $g^{-\epsilon}$. Then [its first derivative](https://youtu.be/mV0N3qHc4Bo?t=318) is $\color{limegreen}{D_x \\; g^{-\epsilon} = \epsilon g^{-1-\epsilon} g'}$. Then $g^{-\epsilon} > 0$ and tends to 0, and ${\color{limegreen}{D_x \\; g^{-\epsilon}}} < 0 $.
If ${\color{limegreen}{D_x \\; g^{-\epsilon}}} < -\epsilon$ on a set of infinite measure, then $\int^{x}_0 {\color{limegreen}{D_x \, g^{-\epsilon}}} \quad dx $ (which differs from $g^{-\epsilon}$ by a constant) would diverge to negative infinity. So ${\color{limegreen}{D_x \\; g^{-\epsilon}}} \ge -\epsilon$ outside a set of finite measure, from which your desired inequality follows immediately.