# Show that $m\le\left(\frac{1}{b-a}\int_{a}^{b}f^{2}\left(x\right)dx\right)^{\frac{1}{2}}\le M$

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.

Hint:

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 $$|M|$$.

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