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Is the following true :

Let $f$ be a positive continuous function such that : $\int_{-\infty}^{+\infty} f(t) \mathrm{d}t = 1$, and such that : $\int_{- \infty}^{+\infty} tf(t) \mathrm{d}t < \infty$. If $\phi$ is a continuous function and : $\int_{-\infty}^{+\infty} t\phi(f(t)) \mathrm{d}t < \infty$ then do we have : $$\int_{-\infty}^{+\infty} t \phi(f(t)) \mathrm{d}t = \int_{-\infty}^{+\infty} \phi(t) f(t) \mathrm{d}t$$ ?

I think there is a link with probability, yet I don't know if this is true. It kind of reminds me expectation. For example in the discrete case with a random variable $X$ we have : $\mathbb{E}[\phi(X)] = \sum \phi(x) \mathbb{P}(X =x)$ yet I don't know if it extends to the continous case.

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2 Answers 2

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Take $\phi (t)=t^{2}$. The equation becomes $\int tf^{2}(t)dt=\int t^{2}f(t)dt$ and the left side is $0$ whenever $f$ is an even function. The right side is positive whenever $f$ is positive, so $f(t)=ce^{-t^{2}}$ with a suitable $c>0$ gives a counterexample.

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The answer seems to be no in general.

Take $f(t) = \frac{C}{t}$, $\phi(t)=t^2$ on some finite domain $D$ (like $[1;2]$), with $C = 1/\int \frac{1}{t}$.

Left handside gives $\int_D C^2/t = C$ and right handside of the equation gives $\int_D C\times t$ which seems hard to be true for any domain $D$.

Continuity can be achieved with an arbitrary small error.

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