Counterexample of the converse of Jensen's inequality

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!

• 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. – Ian Nov 15 '18 at 1:39
• @Ian so you mean I can use $\phi=\Big|\dfrac{1}{x}\Big|$ and $f(x)=x$ but just to restrict the case to $[0,1]$? – JacobsonRadical Nov 15 '18 at 1:42
• 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. – Ian Nov 15 '18 at 2:52
• @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 $[0,1]$. – JacobsonRadical Nov 15 '18 at 3:03

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 $$\mathbb{R}$$.