# MVT for integrals: strict inequality not needed before applying IVT?

I've looked at Nigel Overmars's answer here: https://math.stackexchange.com/a/630429/349828

His proof is essentially identical to the one I wrote myself and to the one given by my analysis professor in class. However, I'm driven out of an abundance of caution to ask the following question:

Are we really allowed to apply the Intermediate Value Theorem once we have: $$m \leq \frac{\int_a^b f(x)g(x)dx}{\int_a^b g(x)dx} \leq M$$

I was under the impression that we need strict inequality for the hypotheses of the IVT to be satisfied, i.e., we need the following: $$m \lt \frac{\int_a^b f(x)g(x)dx}{\int_a^b g(x)dx} \lt M$$

Am I crazy?

Edited to add: Here's the full question I'm trying to answer. I'm fairly sure I did part (a) without any trouble. The subtle snag I'm worried about is in part (b), which led to my original question above.

Suppose that $a \lt b$, $f$ and $g$ are continuous real-valued functions on the interval $[a, b]$, and $0 \lt g(x)$ for each $x$ in $[a, b]$. Prove each of the following:

a. If $m \leq f(x) \leq M$ on $[a, b]$ for some real numbers $m$ and $M$, then $m \leq \frac{\int_a^b f(x)g(x)dx}{\int_a^b g(x)dx} \leq M .$

b. There exists $c \in (a, b)$ such that $\int_a^b f(x)g(x)dx = f(c)\int_a^b g(x)dx .$

Does that make my concern any clearer, or am I still crazy?

• Normally the statement of IVT is given with strict inequalities, but the result is true with weak inequalities also but then the conclusion of IVT also deals with weaker inequality. – Paramanand Singh Jun 26 '16 at 8:38
• That makes sense. Thank you! – MMASRP63 Jun 26 '16 at 19:48

From OP's question and answer it is clear that he is interested in ensuring that $c \in (a, b)$ instead of $c \in [a, b]$ in the statement of mean value theorem for integrals. The result is true but somewhat difficult to establish. A much simpler approach is to use the Cauchy's Mean Value theorem with $$F(x) = \int_{a}^{x}f(t)g(t)\,dt, G(x) = \int_{a}^{x}g(t)\,dt$$ Since $f, g$ are continuous on $[a, b]$ and $g$ is positive, the functions $F, G$ satisfy all the conditions of Cauchy's Mean Value Theorem and therefore $$\frac{F(b) - F(a)}{G(b) - G(a)} = \frac{F'(c)}{G'(c)}$$ for some $c \in (a, b)$. This means that $$\int_{a}^{b}f(x)g(x)\,dx = f(c)\int_{a}^{b}g(x)\,dx$$ for some $c \in (a, b)$.

This is one of the reasons that the mean value theorem for derivatives is a much more powerful result than the mean value theorem for integrals. See this answer also in this regard.

• That is brilliant, and very elegant! Thank you so much. I'm studying for a comprehensive exam in real analysis, and this will be very helpful. I've selected your response as the accepted answer to the question instead of mine since it is spot-on. – MMASRP63 Jun 26 '16 at 19:54
• @MMASRP63: wish you best of luck for your exam. – Paramanand Singh Jun 27 '16 at 3:30
• @MMASRP63: also note that this answer uses the Fundamental Theorem of Calculus. If you are not allowed to use this theorem then I think your answer is the only option. This shows the power of FTC also. – Paramanand Singh Jun 27 '16 at 3:39

The relevant function to consider here is $$F(c) = \frac{f(c) \int_a^b g(x) dx}{\int_a^b g(x) dx}.$$ Then $F$ is continuous and so it maps the interval $[a,b]$ to the interval $[m, M]$.

What this means is that there is some $c \in [a,b]$ such that $$F(c) = \frac{\int_a^b f(x) g(x) dx}{\int_a^b g(x)dx}.$$ This is what the problem you link to asked to show.

It is conceivable that $c = a$ or $c = b$, so one can only guarantee $c \in [a,b]$ instead of $c \in (a,b)$.

• Thank you. I see you edited the linked solution to conclude that $c \in [a, b]$ rather than $c \in (a, b)$. This makes sense to me, and I believe it answers the original question in the post to which I linked. However, the question I am trying to answer actually does require showing that $c \in (a, b)$. That's why I'm stuck. I will update my question to make this clearer. Thanks again! – MMASRP63 Jun 23 '16 at 23:18

I believe I have come up with an adequate answer to this question. Here it is:

$f$ and $g$ are given to be continuous on $[a, b]$, so they are both integrable on $[a, b]$. Also, $g(x)$ is given to be positive for all $x \in [a, b]$.

Part (a):

Assume that $m \leq f(x) \leq M$ for some $m, M \in \Bbb R$ and for all $x \in [a, b]$.

Remark: the problem gives this to us by assumption, but in fact, we are guaranteed to have such $m$ and $M$ by the Extreme Value Theorem.

Since $g(x) \gt 0$ for all $x \in [a, b]$, we have: $$mg(x) \leq f(x)g(x) \leq Mg(x) \qquad \text{for all } x \in [a, b]$$ $$\int_a^bmg(x)dx \leq \int_a^bf(x)g(x)dx \leq \int_a^bMg(x)dx$$ $$m\int_a^bg(x)dx \leq \int_a^bf(x)g(x)dx \leq M\int_a^bg(x)dx$$ Since $g(x) \gt 0$ for all $x \in [a, b]$, we know $\int_a^bg(x)dx \gt 0$. Thus, we have: $$m \leq \frac{\int_a^bf(x)g(x)dx}{\int_a^bg(x)dx} \leq M$$as desired. QED.

Part (b):

Let $m = \inf\{f(x)\,\lvert\, x \in [a, b]\}$ and $M = \sup\{f(x)\,\lvert\, x \in [a, b]\}$.

Since $f$ is continuous and $[a, b]$ is compact (by the Heine-Borel Theorem), we are guaranteed by the Extreme Value Theorem that:

(1) $m \in \Bbb R$ and $M \in \Bbb R$

(2) $f(s) = m$ and $f(t) = M$ for some $s, t \in [a, b]$

We must have $m \leq M$, so we will consider two cases: $m = M$ or $m \lt M$.

Case One: Assume that $m = M$. Then $f$ is a constant function on $[a, b]$. Hence, for any $c \in (a, b)$ it is trivially true that $\int_a^bf(x)g(x)dx = f(c)\int_a^bg(x)dx$.

Case Two: Assume that $m \lt M$. Let $I \subseteq \Bbb R$ be the image of $[a, b]$ under $f$, i.e., $I = f([a, b])$. Since $f$ is continuous and $[a, b]$ is connected, $I$ is connected. The only connected subsets of $\Bbb R$ are intervals. Since $m = \inf I$; $M = \sup I$; $m, M \in I$; and $m \neq M$, we must have $I = [m, M]$.

Now, let $J \subseteq \Bbb R$ be the inverse image of $(m, M)$ under $f$, i.e., $J = f^{-1}[(m, M)]$. By the preceding argument, $J$ is non-empty. Since $f$ is continuous and $(m, M)$ is open, $J$ is open. Every non-empty open subset of $\Bbb R$ is a countable union of disjoint open intervals, so this is true of $J$. Let $K \subseteq J$ be one such open interval. Then $m \lt f(x) \lt M$ for all $x \in K$.

We now proceed in a manner similar to part (a) above, but with an important variation.

Since $g(x) \gt 0$ for all $x \in [a, b]$, we have: $$mg(x) \leq f(x)g(x) \leq Mg(x) \qquad \text{for all } x \in [a, b]$$

Now, since $m \lt f(x) \lt M$ for all $x \in K$ and $K \subseteq [a, b]$, we may change to strict inequalities upon integration:

$$\int_a^bmg(x)dx \lt \int_a^bf(x)g(x)dx \lt \int_a^bMg(x)dx$$ $$m\int_a^bg(x)dx \lt \int_a^bf(x)g(x)dx \lt M\int_a^bg(x)dx$$ Since $g(x) \gt 0$ for all $x \in [a, b]$, we know $\int_a^bg(x)dx \gt 0$. Thus, we have: $$m \lt \frac{\int_a^bf(x)g(x)dx}{\int_a^bg(x)dx} \lt M$$ $$f(s) \lt \frac{\int_a^bf(x)g(x)dx}{\int_a^bg(x)dx} \lt f(t)$$

Finally, since $f$ is continuous and $[a, b]$ is connected, we are guaranteed by the Intermediate Value Theorem that there is some $c$ between $s$ and $t$ such that: $$f(c) = \frac{\int_a^bf(x)g(x)dx}{\int_a^bg(x)dx}$$

Therefore, there is indeed a $c \in (a, b)$ such that $\int_a^bf(x)g(x)dx = f(c)\int_a^bg(x)dx$.

In both cases, we were able to supply the desired $c$. QED.

• I think you have done over-complication in part (b). Note that if $f$ is continuous on $[a, b]$ then image of $[a, b]$ under $f$ is a closed interval $[c, d]$ where $c = \inf\, f(x), d = \sup\,f(x)$. And there is no need to go for strict inequalities. It does not give you anything extra. – Paramanand Singh Jun 26 '16 at 8:20
• I agree; I felt I was beating that part of the question to death, but I also think that writing everything out in excessive detail helped me to better understand the mechanics of what was going on. However, I don't understand why you say strict inequalities aren't required. Don't we need strict inequalities if we want to ensure that $c \in (a, b)$ rather than just $c \in [a, b]\,$? Thanks again for all your help and feedback. – MMASRP63 Jun 26 '16 at 20:00
• I read your answer again and it is correct. You do need strict inequality. +1 from my end. – Paramanand Singh Jun 27 '16 at 3:27