Interchange Integration and Minimization

Question

I am new to calculus of variations and I have a problem of the type \begin{equation*} \text{min}_{g\in L^2(\mathbb{R})} \int_{\mathbb{R}} f(x,g(x))\, \mathrm dx, \end{equation*} where $f$ is absolutely integrable and $f\geq 0$. I am able to solve this problem using a classical variational approach. However, as $f\geq 0$ I am wondering whether it is posible to interchange the minimum and the integral and solve \begin{equation*} \text{min}_{g\in L^2(\mathbb{R})}\,\, f(x,g(x)) \quad \forall x\in\mathbb{R} \end{equation*} instead. I would be very happy if someone has any ideas on how to prove/falsify this! Thanks a lot!

Answer 1

The problem is that $$ \min_{g \in L^2(\mathbb{R})}f(x,g(x)) $$ is rather "ill-defined". It can only be taken in the point-wise sense, which is to say that there exists a $\bar{g} \in L^2(\mathbb{R})$ s.t. for all $g \in L^2(\mathbb{R})$ we have $$ f(g(x),x) \geq f(\bar{g}(x),x) \; \forall x \in \mathbb{R} $$ This pointwise condition is way too strong and is indeed fully sufficient:
We could just use monoticity of the integral w.r.t. its integrand and see that $$ \int_{\mathbb{R}} f(g(x),x)dx \geq \int_{\mathbb{R}} f(\bar{g}(x),x)dx $$ And by definition, it would be the minimum. While this could be useful in certain situations for specific choices of $f(p,x)$, it is not useful in general as the point-wise minimization problem is much harder to solve than the one involving the integral.
As you already mentioned, Hilbert space/functional analysis methods (or the direct method, as it is often called) will provide you with an existence results way faster.