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If $f$ is $beta$-smooth and non-negative, then $|f'(x)|^2le 2beta f(x)$?
1
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I am reading a paper, and I found this conclusion from a proof.
I am wondering why we can conclude that if a function $f$ is $beta$-smooth and non-negative, then $|f'(x)|^2le 2beta f(x)$.
A simple definition of $beta$-smooth is $|f''(x)| le beta$.
real-analysis functional-analysis analysis functions optimization
asked Jan 19 at 0:54
coolcatcoolcat
1109
$endgroup$
add a comment |
1
$begingroup$
I am reading a paper, and I found this conclusion from a proof.
I am wondering why we can conclude that if a function $f$ is $beta$-smooth and non-negative, then $|f'(x)|^2le 2beta f(x)$.
A simple definition of $beta$-smooth is $|f''(x)| le beta$.
real-analysis functional-analysis analysis functions optimization
asked Jan 19 at 0:54
coolcatcoolcat
1109
$endgroup$
1
$begingroup$
If additionally you assume that $f'(x)ge 0$ everywhere and for some $x_0$, we have $f(x_0)=0=f'(x_0)$, then $f''(x)≤beta$ implies $$ f''(x)f'(x) ≤ beta f'(x).$$ This LHS is $((f')^2)'/2$. Integrating from $x_0$ to $x$ gives $$ f'(x)^2 le 2beta f(x).$$
$endgroup$
– Calvin Khor
Jan 19 at 21:04
$begingroup$
can you share the paper where this is from?
$endgroup$
– Calvin Khor
Jan 22 at 16:35
1
$begingroup$
Thank you Khor. Your answer provides a good point to me. The conclusion comes from the page 4 (top) of arxiv.org/pdf/1811.02564.pdf.
$endgroup$
– coolcat
Jan 22 at 18:09
add a comment |
1
1
1
1
$begingroup$
I am reading a paper, and I found this conclusion from a proof.
I am wondering why we can conclude that if a function $f$ is $beta$-smooth and non-negative, then $|f'(x)|^2le 2beta f(x)$.
A simple definition of $beta$-smooth is $|f''(x)| le beta$.
real-analysis functional-analysis analysis functions optimization
asked Jan 19 at 0:54
coolcatcoolcat
1109
$endgroup$
I am reading a paper, and I found this conclusion from a proof.
I am wondering why we can conclude that if a function $f$ is $beta$-smooth and non-negative, then $|f'(x)|^2le 2beta f(x)$.
A simple definition of $beta$-smooth is $|f''(x)| le beta$.
real-analysis functional-analysis analysis functions optimization
real-analysis functional-analysis analysis functions optimization
asked Jan 19 at 0:54
coolcatcoolcat
1109
asked Jan 19 at 0:54
coolcatcoolcat
1109
edited Jan 19 at 20:29
coolcat
asked Jan 19 at 0:54
coolcatcoolcat
1109
asked Jan 19 at 0:54
coolcatcoolcat
1109
asked Jan 19 at 0:54
coolcatcoolcat
1109
1109
1
$begingroup$
If additionally you assume that $f'(x)ge 0$ everywhere and for some $x_0$, we have $f(x_0)=0=f'(x_0)$, then $f''(x)≤beta$ implies $$ f''(x)f'(x) ≤ beta f'(x).$$ This LHS is $((f')^2)'/2$. Integrating from $x_0$ to $x$ gives $$ f'(x)^2 le 2beta f(x).$$
$endgroup$
– Calvin Khor
Jan 19 at 21:04
$begingroup$
can you share the paper where this is from?
$endgroup$
– Calvin Khor
Jan 22 at 16:35
1
$begingroup$
Thank you Khor. Your answer provides a good point to me. The conclusion comes from the page 4 (top) of arxiv.org/pdf/1811.02564.pdf.
$endgroup$
– coolcat
Jan 22 at 18:09
add a comment |
1
$begingroup$
If additionally you assume that $f'(x)ge 0$ everywhere and for some $x_0$, we have $f(x_0)=0=f'(x_0)$, then $f''(x)≤beta$ implies $$ f''(x)f'(x) ≤ beta f'(x).$$ This LHS is $((f')^2)'/2$. Integrating from $x_0$ to $x$ gives $$ f'(x)^2 le 2beta f(x).$$
$endgroup$
– Calvin Khor
Jan 19 at 21:04
$begingroup$
can you share the paper where this is from?
$endgroup$
– Calvin Khor
Jan 22 at 16:35
1
$begingroup$
Thank you Khor. Your answer provides a good point to me. The conclusion comes from the page 4 (top) of arxiv.org/pdf/1811.02564.pdf.
$endgroup$
– coolcat
Jan 22 at 18:09
1
1
$begingroup$
If additionally you assume that $f'(x)ge 0$ everywhere and for some $x_0$, we have $f(x_0)=0=f'(x_0)$, then $f''(x)≤beta$ implies $$ f''(x)f'(x) ≤ beta f'(x).$$ This LHS is $((f')^2)'/2$. Integrating from $x_0$ to $x$ gives $$ f'(x)^2 le 2beta f(x).$$
$endgroup$
– Calvin Khor
Jan 19 at 21:04
$begingroup$
If additionally you assume that $f'(x)ge 0$ everywhere and for some $x_0$, we have $f(x_0)=0=f'(x_0)$, then $f''(x)≤beta$ implies $$ f''(x)f'(x) ≤ beta f'(x).$$ This LHS is $((f')^2)'/2$. Integrating from $x_0$ to $x$ gives $$ f'(x)^2 le 2beta f(x).$$
$endgroup$
– Calvin Khor
Jan 19 at 21:04
$begingroup$
can you share the paper where this is from?
$endgroup$
– Calvin Khor
Jan 22 at 16:35
$begingroup$
can you share the paper where this is from?
$endgroup$
– Calvin Khor
Jan 22 at 16:35
1
1
$begingroup$
Thank you Khor. Your answer provides a good point to me. The conclusion comes from the page 4 (top) of arxiv.org/pdf/1811.02564.pdf.
$endgroup$
– coolcat
Jan 22 at 18:09
$begingroup$
Thank you Khor. Your answer provides a good point to me. The conclusion comes from the page 4 (top) of arxiv.org/pdf/1811.02564.pdf.
$endgroup$
– coolcat
Jan 22 at 18:09
add a comment |
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1
$begingroup$
If additionally you assume that $f'(x)ge 0$ everywhere and for some $x_0$, we have $f(x_0)=0=f'(x_0)$, then $f''(x)≤beta$ implies $$ f''(x)f'(x) ≤ beta f'(x).$$ This LHS is $((f')^2)'/2$. Integrating from $x_0$ to $x$ gives $$ f'(x)^2 le 2beta f(x).$$
$endgroup$
– Calvin Khor
Jan 19 at 21:04
$begingroup$
can you share the paper where this is from?
$endgroup$
– Calvin Khor
Jan 22 at 16:35
1
$begingroup$
Thank you Khor. Your answer provides a good point to me. The conclusion comes from the page 4 (top) of arxiv.org/pdf/1811.02564.pdf.
$endgroup$
– coolcat
Jan 22 at 18:09