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Bounding the absolute value of a function with an integral

Prove that the Riemann integral of this function is zeroThe absolute value of a Riemann integrable function is Riemann integrable.Problem on using the definition of Riemann IntegralWhat is $int f$ if $f$ is not Riemann integrable in the reverse direction of this theoremEstimating the Riemann integral of $f$ using an upper bound for $f$Limit of an integrable function on [0,1]Why was it necessary for the Riemann integral to consider all partitions and taggings?Help understanding the proof that a Riemann Integrable function is boundedHelp in understanding a boundedness proof for Riemann integrable functionsHow to calculate the upper and lower Riemann sums with respect to this particular partition P?

6

1

$begingroup$

I am having trouble with the following problem in analysis:

Suppose that $f, f^prime in C([0, 1])$. Prove that for all $x in [0, 1]$
$$
|f(x)| leq int_0^1 (|f(t)| + |f^prime (t)|) dt.
$$

Any pointers? I have tried writing this as a Riemann Sum (given arbitrary tagged partition) but am still not sure how to proceed.

real-analysis

share|cite|improve this question

edited May 6 at 3:57

onesix

asked May 6 at 3:24

onesixonesix

616

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The constant function $f(t) = 1$ is a counterexample to the stronger inequality, taking any $x < 1$.
  $endgroup$
  – Theo Bendit
  May 6 at 3:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, I did not realize this. Should have examined it a little more closely, thanks.
  $endgroup$
  – onesix
  May 6 at 3:50
  

  
  
  
  

add a comment | 

6

1

$begingroup$

I am having trouble with the following problem in analysis:

Suppose that $f, f^prime in C([0, 1])$. Prove that for all $x in [0, 1]$
$$
|f(x)| leq int_0^1 (|f(t)| + |f^prime (t)|) dt.
$$

Any pointers? I have tried writing this as a Riemann Sum (given arbitrary tagged partition) but am still not sure how to proceed.

real-analysis

share|cite|improve this question

edited May 6 at 3:57

onesix

asked May 6 at 3:24

onesixonesix

616

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The constant function $f(t) = 1$ is a counterexample to the stronger inequality, taking any $x < 1$.
  $endgroup$
  – Theo Bendit
  May 6 at 3:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, I did not realize this. Should have examined it a little more closely, thanks.
  $endgroup$
  – onesix
  May 6 at 3:50
  

  
  
  
  

add a comment | 

$begingroup$

I am having trouble with the following problem in analysis:

Suppose that $f, f^prime in C([0, 1])$. Prove that for all $x in [0, 1]$
$$
|f(x)| leq int_0^1 (|f(t)| + |f^prime (t)|) dt.
$$

Any pointers? I have tried writing this as a Riemann Sum (given arbitrary tagged partition) but am still not sure how to proceed.

real-analysis

share|cite|improve this question

edited May 6 at 3:57

onesix

asked May 6 at 3:24

onesixonesix

616

$endgroup$

share|cite|improve this question

edited May 6 at 3:57

onesix

asked May 6 at 3:24

onesixonesix

616

share|cite|improve this question

edited May 6 at 3:57

onesix

asked May 6 at 3:24

onesixonesix

616

asked May 6 at 3:24

onesixonesix

616

asked May 6 at 3:24

onesixonesix

616

1 Answer
1

active

oldest

votes

8

$begingroup$

$int_0^1|f(t)| dt$ is the average of $|f(t)|$ in the interval, and $int_0^1|f'(t)|dt$ is the total variation of $f(t)$, if you think about this, it makes sense. By MVT, you have that $$int_0^1|f(t)| dt=|f(a)|$$ for some $ain[0,1]$. Let $xin[0,1]$, WLOG say $x>a$, then beginalign*|f(x)|&leq |f(a)|+|f(x)-f(a)|\
&=int_0^1|f(t)|dt+|int_a^xf'(t)dt|\
&leq int_0^1|f(t)|dt+int_a^x|f'(t)|dt\
&leqint_0^1|f(t)|dt+int_0^1|f'(t)|dt
endalign*

share|cite|improve this answer

answered May 6 at 4:59

Julian MejiaJulian Mejia

1,643212

$endgroup$

add a comment | 

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8

$begingroup$

$int_0^1|f(t)| dt$ is the average of $|f(t)|$ in the interval, and $int_0^1|f'(t)|dt$ is the total variation of $f(t)$, if you think about this, it makes sense. By MVT, you have that $$int_0^1|f(t)| dt=|f(a)|$$ for some $ain[0,1]$. Let $xin[0,1]$, WLOG say $x>a$, then beginalign*|f(x)|&leq |f(a)|+|f(x)-f(a)|\
&=int_0^1|f(t)|dt+|int_a^xf'(t)dt|\
&leq int_0^1|f(t)|dt+int_a^x|f'(t)|dt\
&leqint_0^1|f(t)|dt+int_0^1|f'(t)|dt
endalign*

share|cite|improve this answer

answered May 6 at 4:59

Julian MejiaJulian Mejia

1,643212

$endgroup$

add a comment | 

8

$begingroup$

$int_0^1|f(t)| dt$ is the average of $|f(t)|$ in the interval, and $int_0^1|f'(t)|dt$ is the total variation of $f(t)$, if you think about this, it makes sense. By MVT, you have that $$int_0^1|f(t)| dt=|f(a)|$$ for some $ain[0,1]$. Let $xin[0,1]$, WLOG say $x>a$, then beginalign*|f(x)|&leq |f(a)|+|f(x)-f(a)|\
&=int_0^1|f(t)|dt+|int_a^xf'(t)dt|\
&leq int_0^1|f(t)|dt+int_a^x|f'(t)|dt\
&leqint_0^1|f(t)|dt+int_0^1|f'(t)|dt
endalign*

share|cite|improve this answer

answered May 6 at 4:59

Julian MejiaJulian Mejia

1,643212

$endgroup$

add a comment | 

$begingroup$

$int_0^1|f(t)| dt$ is the average of $|f(t)|$ in the interval, and $int_0^1|f'(t)|dt$ is the total variation of $f(t)$, if you think about this, it makes sense. By MVT, you have that $$int_0^1|f(t)| dt=|f(a)|$$ for some $ain[0,1]$. Let $xin[0,1]$, WLOG say $x>a$, then beginalign*|f(x)|&leq |f(a)|+|f(x)-f(a)|\
&=int_0^1|f(t)|dt+|int_a^xf'(t)dt|\
&leq int_0^1|f(t)|dt+int_a^x|f'(t)|dt\
&leqint_0^1|f(t)|dt+int_0^1|f'(t)|dt
endalign*

share|cite|improve this answer

answered May 6 at 4:59

Julian MejiaJulian Mejia

1,643212

$endgroup$

share|cite|improve this answer

answered May 6 at 4:59

Julian MejiaJulian Mejia

1,643212

answered May 6 at 4:59

Julian MejiaJulian Mejia

1,643212

answered May 6 at 4:59

Julian MejiaJulian Mejia

1,643212