# A integral inequality and generalization

If $$f(x)$$ is decreasing then $$\int_{a-1}^{b}f(x)dx\ge \sum_{k=a}^b f(k)\ge \int_a^{b+1}f(x)dx$$.

And reverse for increasing functions.

I try to generalize it

Generalization:

Let $$f(x)$$ be a monotone decreasing function on $$\left(0,+\infty\right)$$. Prove that: $$\int_{a}^{b+1}f(x)dx\leq \sum_{k=a}^{b}f(k)\leq \int_{a-1}^{b}f(x)dx\leq \sum_{k=a-1}^{b-1}f(k)\quad a,b\in\mathbb N$$ When does equality hold? *Note that $$0\notin \mathbb N$$

I have 3 questions as follow :

1) Could you please prove generalization

2) Is this true ? Generalization (inversion) Let $$f(x)$$ be a monotone increasing function on $$\left(0,+\infty\right)$$. Prove that: $$\int_{a}^{b+1}f(x)dx\geq \sum_{k=a}^{b}f(k)\geq \int_{a-1}^{b}f(x)dx\geq \sum_{k=a-1}^{b-1}f(k)\quad a,b\in\mathbb N$$ When does equality hold?

3) Can we have a stronger generalization ?

• There might be some typos in your question: in line 1 the upper index of the sum should be $b$ i guess. In the other two inequalities the bounds of the second integral cannot be correct. Please edit!
– weee
Oct 30 '18 at 7:01
• I edited my post !
– nam
Oct 30 '18 at 11:37

1) Once you have the fact $$\int_{a-1}^{b}f(x)dx\ge \sum_{k=a}^b f(k)\ge \int_a^{b+1}f(x)dx,\ \ \forall a,b\in \mathbb N,$$ there is not much to prove, for at the expression $$\int_{a}^{b+1}f(x)dx\leq \sum_{k=a}^{b}f(k)\leq \int_{a-1}^{b}f(x)dx\leq \sum_{k=a-1}^{b-1}f(k),\ \forall a,b\in\mathbb N,$$ the first two inequalities follow directly from the fact and the third one follows from the fact applied to $$a-1$$ and $$b-1$$ in the place of $$a$$ and $$b$$, respectively.

2) The proof for increasing functions is analog to that for decreasing functions. Using the fact for increasing functions $$\int_{a-1}^{b}f(x)dx\le \sum_{k=a}^b f(k)\le \int_a^{b+1}f(x)dx,\ \ \forall a,b\in \mathbb N,$$ at the expression $$\int_{a}^{b+1}f(x)dx\geq \sum_{k=a}^{b}f(k)\geq \int_{a-1}^{b}f(x)dx\geq \sum_{k=a-1}^{b-1}f(k),\ \forall a,b\in\mathbb N,$$ the first two inequalities follow directly from the fact and the third one follows from the fact applied to $$a-1$$ and $$b-1$$ in the place of $$a$$ and $$b$$, respectively.

3) I don't know and I can't think of any suggestions.

About the question of when the equality holds for (1), I have an idea. Note that $$\displaystyle\int_{a-1}^{b}f(x)dx = \int_a^{b+1}f(x)dx$$ if, and only if, $$\int_{a-1}^{a}f(x)dx + \int_{a}^{b}f(x)dx = \int_a^{b}f(x)dx + \int_{b}^{b+1}f(x)dx\ \ \Leftrightarrow\ \ \int_{a-1}^{a}f(x)dx = \int_{b}^{b+1}f(x)dx.$$ So our problem is equivalent to analyze which conditions imply that $$\int_{a-1}^{a}f(x)dx = \int_{b}^{b+1}f(x)dx.$$ The simplest condition that guarantees the equality above is:

Condition 1. $$a-1=b$$;

Other two conditions for which the equality immediately holds are:

Condition 2. $$a-1 and $$f$$ is a constant at the interval $$(a-1,b+1)$$;

Condition 3. $$a-1>b$$ and $$f$$ is a constant at the interval $$(b,a)$$;

I guess these are the onliest cases for that the equality holds, the proof is not so difficult.

• To make it all correct you have to impose that $a\leq b$ otherwise the sums are possibly empty, whereas the integrals are possibly negative. Moreover for the second sum we need $a>1$ because $f(0)$ is not defined.
– weee
Oct 30 '18 at 20:36
• Can't we use the algebraic rule $\displaystyle\int_b^a f(x)dx=-\int_a^b f(x)dx$? Oct 30 '18 at 21:04

For the second inequality note that $$\int_{a-1}^b f(x) dx = \sum_{k=a-1}^{b-1} \int_k^{k+1} f(x) dx$$. Now by monotonicity we get that $$\int_k^{k+1} f(x) dx \geq f(k+1)$$ and hence the second inequality. For the third inequality we can use the same reasoning as above except that we use $$\int_k^{k+1} f(x) dx \leq f(k)$$.

As stated above the equality in the first inequality holds if we have a piecewise constant function which is constant on $$[k, k+1)$$ for $$k = a, \dots, b$$. Now assume that we have a function $$g$$ that satisfies the equality. Then $$g$$ is piecewise constant on $$[k, k+1)$$ for $$k = a, \dots, b$$. If it was not then there would exist a $$k \in \{a, \dots, b\}$$ and some $$\xi \in [k, k+1)$$ for which we have $$g(k) > g(\xi)$$ since $$g$$ is decreasing. This would however imply that $$\int_k^{k+1} g(x) dx < g(k)$$ and hence we do not have equality.