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

According to the wikipedia page, the function $\varphi$ must be convex.

I would like to define a function $\varphi(x)$ where $$\varphi(f(x)) = f(x)*h(x)$$

This is because the integral at hand is $\int_0^1 f(x)h(x) \,dx$ and I want to use Jensen's inequality to pull the $h(x)$ outside of the integral, while leaving the $f(x)$ inside.

I can show that $\varphi(x) = x*h(x)$ is convex over $x$. However, I know that's not exactly what I'm doing in the above equation. The first equation is more like $\varphi(t) = t*h(x)$, which I don't think I can say if it's in general convex, since there are two variables (and do I need to show it's convex over $t$? or over $x$?). However, the function is convex over $x$ when I put in the value needed for the integral at hand, which is $f(x)$.

I do not want to use $\varphi(x) = f(x)*h(x)$ because that is not convex, and because I want to keep the $f(x)$ on the inside of the integral.

Can anyone help me understand this issue of convexity of $\phi$ better, and if the function is ok the way I defined it?

Thank you so much!

share|cite|improve this question
The one-loop phi $\varphi$ is called \varphi in TeX, it that's what you want. – Henning Makholm Sep 28 '11 at 20:45
@Henning: thanks, I just changed it! – Angada Sep 28 '11 at 20:48
up vote 1 down vote accepted

In general, $\varphi(f(x)) = f(x) h(x)$ is impossible, because the right side depends on $h(x)$ as well as $f(x)$. The bounds you can get on $\int_0^1 f(x) h(x)\ dx$ (from Hölder, not Jensen) are $ \left| \int_0^1 f(x) h(x)\ dx \right| \le \|f\|_p \|h\|_q$ where $1 \le p,q \le \infty$ and $1/p + 1/q = 1$. Here $\|f\|_p = \left( \int_0^1 |f(x)|^p \, dx\right)^{1/p}$ for $p < \infty$ while $\|f\|_\infty$ is the essential supremum of $|f(x)|$ on $[0,1]$.

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.