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Show that $|f(x)frac{b-a}{(x-a)(x-b)}|leq int_a^b|f''(x)|dx$.

Topological degreeProve $|int_a^b$$f(x)dx| leq int_a^b$$|f(x)|dx$Real Analysis: Show that g is integrable on [a,b] and that $int_a^b$ $g(x)dx=$ $int_a^b$ $f(x)dx$Prove that $int_a^b f(x) , dx >0$$|g(x)| leq K int_a^x|g| forall x in I$Show that if $g:[a,b] tomathbb{R}$ is continuous, then there exists a point $bar{x}in(a,b)$ such that $g(x)=frac{1}{b-a}int_a^b g(t)dt$Proof: $V_a^b(f) = int_a^b |f'(x)|dx$Show that $int_a^b {f+g}=int_a^b f +int_a^b g$Help checking the proof that $int_a^bf(x)>0$ for a conti. function that a point greater than $0$No non-negative continuous function on $[a,b]$ such that $int_a^b f(t)dt=1, int_a^b tf(t)dt=c, int_a^b t^2f(t)dt=c^2$ for $cinmathbb{R}$.

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I need help with the following exercice:

Suppose that $fin C^2[a,b]$, and $f(a)=f(b)=0$. Show that
$|f(x)frac{b-a}{(x-a)(x-b)}|leq int_a^b|f''(x)|dx$.

The best I can get is the following, which is not enough: choose an $xin [a,(a+b)/2]$, then there exists $xiin (a,x)$ such that
$|f(x)frac{b-a}{(x-a)(x-b)}|=|f'(xi)frac{b-a}{x-b}|$. Since $f(a)=f(b)=0$, we know that there exists $etain[a,b]$ such that $f'(eta)=0$. And therefore for some $lambda$ between $xi$ and $eta$, $|f'(xi)frac{b-a}{x-b}|=|f''(lambda)(xi-eta)frac{b-a}{x-b}|leq |2f''(lambda)(xi-eta)|leq 2|f''(lambda)|(b-a)$.

calculus analysis

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asked Mar 12 at 4:34

JiuJiu

520113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont know if it helps, but $f''(x)$ is continuous on $[a,b]$, then attains a minimum, that is, there is $m in [a,b]$ such that $f''(m) leq f''(x)$ for any $x in [a,b]$. So, $|f''(m)|(b-a) = int_{a}^{b}|f''(m)|dx leq int_{a}^{b}|f''(x)|dx$.
  $endgroup$
  – Lucas Corrêa
  Mar 12 at 4:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should try Taylor's theorem with the integral form of the remainder
  $endgroup$
  – AlexL
  Mar 12 at 4:52
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I need help with the following exercice:

Suppose that $fin C^2[a,b]$, and $f(a)=f(b)=0$. Show that
$|f(x)frac{b-a}{(x-a)(x-b)}|leq int_a^b|f''(x)|dx$.

The best I can get is the following, which is not enough: choose an $xin [a,(a+b)/2]$, then there exists $xiin (a,x)$ such that
$|f(x)frac{b-a}{(x-a)(x-b)}|=|f'(xi)frac{b-a}{x-b}|$. Since $f(a)=f(b)=0$, we know that there exists $etain[a,b]$ such that $f'(eta)=0$. And therefore for some $lambda$ between $xi$ and $eta$, $|f'(xi)frac{b-a}{x-b}|=|f''(lambda)(xi-eta)frac{b-a}{x-b}|leq |2f''(lambda)(xi-eta)|leq 2|f''(lambda)|(b-a)$.

calculus analysis

share|cite|improve this question

asked Mar 12 at 4:34

JiuJiu

520113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont know if it helps, but $f''(x)$ is continuous on $[a,b]$, then attains a minimum, that is, there is $m in [a,b]$ such that $f''(m) leq f''(x)$ for any $x in [a,b]$. So, $|f''(m)|(b-a) = int_{a}^{b}|f''(m)|dx leq int_{a}^{b}|f''(x)|dx$.
  $endgroup$
  – Lucas Corrêa
  Mar 12 at 4:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should try Taylor's theorem with the integral form of the remainder
  $endgroup$
  – AlexL
  Mar 12 at 4:52
  

  
  
  
  

add a comment | 

$begingroup$

I need help with the following exercice:

Suppose that $fin C^2[a,b]$, and $f(a)=f(b)=0$. Show that
$|f(x)frac{b-a}{(x-a)(x-b)}|leq int_a^b|f''(x)|dx$.

The best I can get is the following, which is not enough: choose an $xin [a,(a+b)/2]$, then there exists $xiin (a,x)$ such that
$|f(x)frac{b-a}{(x-a)(x-b)}|=|f'(xi)frac{b-a}{x-b}|$. Since $f(a)=f(b)=0$, we know that there exists $etain[a,b]$ such that $f'(eta)=0$. And therefore for some $lambda$ between $xi$ and $eta$, $|f'(xi)frac{b-a}{x-b}|=|f''(lambda)(xi-eta)frac{b-a}{x-b}|leq |2f''(lambda)(xi-eta)|leq 2|f''(lambda)|(b-a)$.

calculus analysis

share|cite|improve this question

asked Mar 12 at 4:34

JiuJiu

520113

$endgroup$

share|cite|improve this question

asked Mar 12 at 4:34

JiuJiu

520113

share|cite|improve this question

asked Mar 12 at 4:34

JiuJiu

520113

asked Mar 12 at 4:34

JiuJiu

520113

asked Mar 12 at 4:34

JiuJiu

520113