Assume $f$ is a real-valued function which is integrable over the interval $I=[a,b]$ and for every $x \in [a,b]$ we have that $0\le m \le f(x) \le M$,show the following inequality does hold: $$m\le\left(\frac{1}{b-a}\int_{a}^{b}f^{2}\left(x\right)dx\right)^{\frac{1}{2}}\le M$$

I know that for $M_I=\sup \{f(x):x \in [a,b]\}$ and $m_I=\inf \{f(x):x \in [a,b]\}$ we have that :

$$ U(P,f)=\sum_{i=0}^{n-1} M_i \Delta x_{i}\le M_I\sum_{i=0}^{n-1} \Delta x_{i} =M_I(b-a)$$

And $$ m_I(b-a)=m_I\sum_{i=0}^{n-1} \Delta x_{i} \le \sum_{i=0}^{n-1} m_i \Delta x_{i}=L(P,f)$$

Where $P=(a=x_0,x_1,...,x_n=b)$ is a partition of $[a,b]$,combining these two inequality gives:

$$m_I(b-a) \le L(P,f) \le U(P,f)\le M_I(b-a)$$

And if the function $f$ is integrable over $[a,b]$,then $$m_I(b-a) \le \int_{a}^{b}f\left(x\right)dx\le M_I(b-a)$$

And if for every $x \in [a,b]$ we have that $0\le m \le f(x) \le M$ then from the definition of supremum and infimum:

$$m \le \frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx\le M$$

But I don't know how the get the main inequality.



If you take the square, what you want to show is $$ m^2\le\frac{1}{b-a}\int_a^b f^2\le M^2 $$

but $m^2\le f^2\le M^2$. Simply integrate it.

You should assume that $m,M\ge 0$. Otherwise, you need to replace them with $|m|$ and $|M|$.

  • $\begingroup$ Why $ m^2 \le f^2 \le M^2$? $\endgroup$ – masaheb Jan 16 at 16:21
  • $\begingroup$ @masaheb because $m\le f\le M$. $\endgroup$ – user9464 Jan 16 at 16:22
  • $\begingroup$ Well why $M \le M^2$? what if $M$ is between $0$ and $1$? $\endgroup$ – masaheb Jan 16 at 16:23
  • $\begingroup$ @masaheb: no, you don't need $M\le M^2$ and I didn't say that. $\endgroup$ – user9464 Jan 16 at 16:23
  • $\begingroup$ So why that's true? $\endgroup$ – masaheb Jan 16 at 16:24

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