Find the minimum of integral

Question

For $g(x):\Bbb R \rightarrow [0,\infty],$ if $$\int_{-\infty}^{\infty} g(x)\, dx= \int_{-\infty}^{\infty} x^2g(x)\, dx=C$$ find the minimum value of $\int_{-\infty}^{\infty} \ln(g(x))g(x)\, dx$.

Would you give me some hint to this?

Thank you.

Answer 1

We want to minimize the functional $$ S[g]=\int_{\mathbb R} g(x)\ln(g(x))\, dx \tag{1} $$ subject to the constraints $$ \int_{\mathbb R} g(x)\, dx=C,\qquad \int_{\mathbb R} x^2g(x)\, dx=C. \tag{2} $$ Using the method of Lagrange multipliers, this is equivalent to finding the stationary point(s) of the modified functional $$ \tilde{S}[g,\lambda,\mu]=\int_{\mathbb R} g(x)\ln(g(x))\,dx+\lambda\left(\int_{\mathbb R} g(x)\, dx-C\right)+\mu\left(\int_{\mathbb R} x^2g(x)\, dx-C\right). \tag{3} $$ The Euler-Lagrange equation for $\tilde{S}$ is $$ \frac{\partial}{\partial g}\left[g(x)\ln(g(x))+\lambda g(x)+\mu x^2g(x)\right]=0 $$ $$ \implies \ln(g(x))+1+\lambda+\mu x^2=0 $$ $$\implies g(x)=e^{-1-\lambda-\mu x^2}, \tag{4} $$ showing that $g$ is a Gaussian function. It is a straightforward calculation to determine the parameters $\lambda$ and $\mu$ using the constraints $(2)$. The final result is $$ g(x)=\frac{C}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}. \tag{5} $$ Plugging $(5)$ into $(1)$, we obtain, after another straightforward calculation, the minimum value of $S[g]$: $$ S_{\min}=-\frac{C}{2}\left(\ln(2\pi)+1\right). \tag{6} $$