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Let $\phi$ be a convex function on $(-\infty, \infty)$, $f$ a Lebesgue integrable function over $[0,1]$ and $\phi\circ f$ also integrable over $[0,1]$. Then we have:

$$\phi\Big(\int_{0}^{1} f(x)dx\Big)\leq\int_{0}^{1}\Big(\phi\circ f(x)\Big)dx.$$

I am thinking about a counterexample that the converse of this statement. In other words, I am trying to find a $\phi$ which is convex on $\mathbb{R}$, and $f$ is a Lebesgue integral function (on some set), satisfying $\phi\Big(\int_{0}^{1} f(x)dx\Big)\leq\int_{0}^{1}\Big(\phi\circ f(x)\Big)dx$, but $\phi\circ f$ is not Lebesgue integrable (on some set).

I tried to use the convexity of non-integrability of $\dfrac{1}{x}$. However, $\dfrac{1}{x}$ is not convex in the whole $\mathbb{R}$, so I tried to use $\left|\dfrac{1}{x}\right|$. So define $\phi:=\left|\dfrac{1}{x}\right|$.

We know that $f(x)=x$ is Lebesgue integrable, and if we restrict our case to $\mathbb{R^{+}}$, then $\phi\circ f=\dfrac{1}{x}$ which is not Lebesgue integrable.

Then, we have $$\phi\Big(\int_{1}^{2} f(x)dx\Big)=\dfrac{2}{3}<\int_{1}^{2}\Big(\phi\circ f(x)\Big)dx=\log(2).$$

Is my argument correct? I feel that I am kind of in a self-contradiction, or my attempt to show the converse of the statement of Jensen's inequality is wrong from the beginning.

Thank you so much for any ideas!

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  • $\begingroup$ Your argument just re-affirms Jensen's inequality because $\phi \circ f$ is actually integrable on $[1,2]$ in the first place. It's not integrable on $[0,1]$ but in this case you just get that the right side is $\infty$ and Jensen's inequality still holds in this situation. About the only way to break it is to have $\phi \circ f$ be non-integrable for $\infty - \infty$ reasons (so that the inequality simply doesn't make sense), I think. $\endgroup$ – Ian Nov 15 '18 at 1:39
  • $\begingroup$ @Ian so you mean I can use $\phi=\Big|\dfrac{1}{x}\Big|$ and $f(x)=x$ but just to restrict the case to $[0,1]$? $\endgroup$ – JacobsonRadical Nov 15 '18 at 1:42
  • $\begingroup$ If you look at $[0,1]$ then the inequality still holds with the right side being infinity. About the only way to really break it is to find a suitable $\phi,f$ so that the integral of $\phi \circ f$ doesn't exist at all. $\endgroup$ – Ian Nov 15 '18 at 2:52
  • $\begingroup$ @Ian Oh! In fact I needs the inequality holds. I just needs $\phi\circ f$ is not Lebesgue integrable, so as you pointed out, at $[0,1]$ the integral of $\phi\circ f$ is infinity, so $\phi\circ f$ is not Lebesgue integrable on $[0,1]$. $\endgroup$ – JacobsonRadical Nov 15 '18 at 3:03
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Let me restate my question here.

In fact, I need to find a $\phi(x)$ which is convex on whole $\mathbb{R}$, $f(x)\in L^{1}(\mathbb{R})$ and $\phi(\int f)\leq\int\phi(f)$, but $\phi(f)\notin L^{1}(\mathbb{R})$.

In my post before, I used $f(x)=x$, but I realized that $f(x)$ is no integrable over $\mathbb{R}$, so I modified it a little bit and now here is a valid counter-example.

Consider $\phi(x)=\Big|\dfrac{1}{x}\Big|$ and $f(x)=x$ if $x\in[0,1]$ but $f(x)=0$ at all other $x$. It is clear that $\phi(x)$ is convex on $\mathbb{R}$, and $f(x)\in L^{1}(\mathbb{R})$.

Now, $\phi(\int_{0}^{1}f(x)dx)=2$, but $\int_{0}^{1}\phi(f(x))dx=\infty$.

Thus, we have $\phi(\int_{0}^{1}f(x)dx)<\int_{0}^{1}\phi(f(x))dx$ but $\phi(f(x))$ is not Lebesgue integrable over $[0,1]$ by definition, and thus it cannot be integrable over $\mathbb{R}$.

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