# Defining a phi for Jensen's inequality: specifics on convexity

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!

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1
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

## 1 Answer

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 $[0,1]$.

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