Suppose we have the following optimization \begin{align} A=\inf_{g(x) \in D} \left(\int |f(x)-g(x)|^p d\mu \right) \end{align} For $p \ge 1$, and where $\mu$ is a finite measure and $f(x)\in L^p$ and $D$ is some set to which $f(x)$ might or might not belong. Let $\hat{g}(x)$ be the function that attains the minimum
Dose the fact that $A <\infty$ imply that $\hat{g}(x) \in L^p$?
Here is my attempt at a proof \begin{align} A^{1/p}&= \left(\inf_{g(x)} \left(\int |f(x)-g(x)|^p d\mu \right) \right)^{1/p}\\ &=\inf_{g(x)} \left(\int |f(x)-g(x)|^p d\mu \right)^{1/p}\\ &=\inf_{g(x)} || f(x)-g(x)||_p\\ & \ge || f(x)-\hat{g}(x)||_p\\ & \ge \left| ||f(x)||_p-||\hat{g}(x)||_p \right| \text{ where we have used the revers triangle inequality } \end{align}
So, we have that \begin{align} \left| ||f(x)||_p-||\hat{g}(x)||_p \right| \le A^{1/p} \Rightarrow ||\hat{g}(x)||_p \le A^{1/p} +||f(x)||_p \end{align} and therefore $\hat{g}(x) \in L^p$.
Please let me know if my proof is correct and how can I improve it? Thank you
If $g\notin L^p$ then the integral is $+\infty$, because if you assume the contrary, then $f-g\in L^p$, but as $L^p$ is a linear space and $f\in L^p\Rightarrow g\in L^p$. Next, there might be no element $\hat g\in D$ for which the minimum is attained and in this case you do not have something to worry about. But if it is attained, then the integral is finite, which means that $\hat g\in L^p$, because like above $L^p$ is linear, $f\in L^p$ and $f-\hat g\in L^p$