If $f$ is $beta$-smooth and non-negative, then $|f'(x)|^2le 2beta f(x)$?












1












$begingroup$


I am reading a paper, and I found this conclusion from a proof.enter image description here



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







share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


I am reading a paper, and I found this conclusion from a proof.enter image description here



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







share|cite|improve this question











$endgroup$








  • 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














1












1








1


1



$begingroup$


I am reading a paper, and I found this conclusion from a proof.enter image description here



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







share|cite|improve this question











$endgroup$




I am reading a paper, and I found this conclusion from a proof.enter image description here



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 20:29







coolcat

















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














  • 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










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