Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have a function, $\phi:[a,b]\to\mathbb{R}_{+}$. I am trying to prove that the function:


attains its minimum value on $(a,b)$, and find the point in which it reaches that value (I believe it can be eventually expressed in terms of the function $\phi$).

The results I've obtained so far are not very promising. First off, the only way that I know to prove the thesis is by finding an $\alpha_{0}$, such that $g^{\prime}(\alpha_{0})=0$ and then show that $g^{\prime\prime}(\alpha_{0})\geq{0}$.

As for the first part, let's define a bivariate function: $h(\alpha,x)=(x-\alpha)^{2}\phi(x)$. We have:


I have no idea how to deal with the expression $\int^{b}_{a}x\phi(x)\,dx$. Is there a way we can somehow evaluate it and transform into a more elementary form? If not, then by equating the result we obtained to $0$, we arrive at:


How can we interpret the above expression, or at least prove that it belongs to the interval $(a,b)$? I would be very thankful on some ideas as to where I should head with this.

EDIT: we also assume that $\phi$ is continuous.

share|cite|improve this question
Are there any smoothness assumptions about $\phi $? – user17794 Sep 29 '12 at 17:27
No, nothing beyond of what I mentioned. – Johnny Westerling Sep 29 '12 at 17:32
Sorry, my bad! We do assume that $\phi$ is continuous. – Johnny Westerling Sep 29 '12 at 17:36
Can't you just work with the sign of the derivative? $\phi$ is >0 on the whole interval, that means its integral is positive... – Juan Sebastian Totero Sep 29 '12 at 17:41
up vote 2 down vote accepted

Since $\phi$ continuos, $c:=\min\{\phi(x)\mid a\le x\le b\}$ is assumed at some point in $[a,b]$. Since $\phi$ is strictly positive, we conclude $c>0$ , we have that $(x-a)\phi(x)\ge c(x-a)$ and $(b-x)\phi(x)\ge c(b-x)$ for all $x\in[a,b]$, hence $$\int_a^b x \phi(x) dx\ge \int_a^b a \phi(x) dx+\int_a^b c(x-a)dx=a\int_a^b\phi(x)dx+\frac c2(b-a)^2.$$ Similarly, $$\int_a^b x \phi(x) dx\le \int_a^b b \phi(x) dx-\int_a^b c(b-x)dx=a\int_a^b\phi(x)dx-\frac c2(b-a)^2.$$ Because $\frac c2(b-a)^2>0$ and $\int_a^b\phi(x)dx>0$ we conclude

$$ a< \frac{\int_a^b x \phi(x) dx}{\int_a^b \phi(x) dx}< b.$$

Remark: Even if we only assume $\phi(x)\ge0$ for $x\in[a,b]$ and only $\phi(x_0)\ne0$ for some $x_0\in [a,b]$, continuity of $\phi$ allows us to find a subinterval $[a',b']$ around $x_0$ where $\phi$ is strictly bigger than the positive number $c':=\frac12\phi(x_0)$. Then we still have strict inequalities because we may replace the expression $\frac c2(b-a)^2$ with $\frac {c'}2(b'-a')^2$ in the above argument.

Once you have established $a<\alpha<b$ this way, you of course have immediately that $g''(\alpha)=2\int_a^b\phi(x)dx>0$, i.e. $g(x)$ takes has a local minimum at $x=\alpha$. This is also the global minimum for $[a,b]$ because a minimum at the boundary (i.e. at $x=a$ or $x=b$) would require a local maximum inbetween.

share|cite|improve this answer
Thanks. Ok, I do understand the first part, but how does the fact that $\int^{b}_{a}\phi(x)dx>0$ bring us to the second part? – Johnny Westerling Sep 29 '12 at 17:48
Thank you very much for a thorough explanation! – Johnny Westerling Sep 29 '12 at 18:24

If the function $\phi$ is continuous, $\alpha$ could be interpreted, from a mechanical point of view, as the center of mass abscissa of a material line (the segment $[a,b]$) of linear mass density $\phi(x)$.

Another interpretation of $\alpha$ is, being $\phi$ positive, and if it is left continuous, as the mean value of the probability density distribution

$$ \psi(x)=\frac{1}{N}\phi(x),\qquad N=\int_a^b\phi(x)dx. $$

Obviously, both interpretations lead to $a\leq \alpha\leq b$, that can be easily proved directly.

share|cite|improve this answer
Oook, I do not understand the mechanical interpretation. I do see how this works in the second case (the mean must lie within the interval on which the pdf is defined), though I would appreciate some hints on how a rigorous proof should look (I mean, if we shift to probability, we have to prove that $\psi$ satisfies conditions to be a pdf) – Johnny Westerling Sep 29 '12 at 17:55
@JohnnyWesterling: continuity and positivity are enough for a PDF. Also, in your question you ask for an interpretation, so I talked about that. The formal proof is given in the answer by HagenvonEitzen. – enzotib Sep 29 '12 at 17:58
And what about the condition that $\int\psi(x)dx=1$? – Johnny Westerling Sep 29 '12 at 18:03
@JohnnyWesterling: it is guaranteed, in this case, by its definition (I mean of $\psi$) and that of $N$. – enzotib Sep 29 '12 at 18:04
Alright, I see it now. Thanks! – Johnny Westerling Sep 29 '12 at 18:27

It is widely known that for random variables $X$ for which the expected value $E|X|$ is finite, the value of $\alpha$ that minimizes $E((X-\alpha)^2)$ is $\alpha=E(X)$.

So in case $\varphi$ is a probability density function, that answers the question. Where $\varphi$ is a non-negative function whose integral is finite, just divide both sides of the inequality by that integral, and again, that answers the question.

Here's a proof: Consider $$ \alpha\mapsto E((X-\alpha)^2) = E(X^2) -2\alpha E(X) + \alpha^2. $$ No you're just minimizing a quadratic polynomial in $\alpha$.

share|cite|improve this answer
Yes, I see it. Thanks! – Johnny Westerling Sep 29 '12 at 20:13

Let $h(\alpha)=g'(\alpha)$. Then $h$ is continuous everywhere (in fact it is a linear function) with $$h(a)=2\int_a^b(a-x)\phi(x)dx<0$$ and $$h(b)=2\int_a^b(b-x)\phi(x)dx>0$$ by positivity of $\phi$. It follows that there is a zero of $g'$ in $(a,b)$, which the second derivative test shows to be a minimum of $g$. By linearity, $g'$ has a unique zero, the location of which (given in terms of integrals of $\phi$) you have already determined.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.