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hthi
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There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that $$ \int_{B_R(0)}\|u\|_pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u). $$$$ \int_{B_R(0)}\|u\|^pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|^pd\mathbb{P}(u). $$ Then if $x > B_{3R}(0)$, we see that $$ \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u) $$$$ \int_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|^pd\mathbb{P}(u) $$ and hence is $$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u). $$$$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u). $$ Now, just note that $x\mapsto \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u)$$x\mapsto \int_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B_{3R}(0)$. Moreover, we can say that the minimizer is unique since the function is convex (that follows from the convexity of the $p$-norm$x\mapsto \|x\|^p$).

There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that $$ \int_{B_R(0)}\|u\|_pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u). $$ Then if $x > B_{3R}(0)$, we see that $$ \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u) $$ and hence is $$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u). $$ Now, just note that $x\mapsto \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B_{3R}(0)$. Moreover, we can say that the minimizer is unique since the function is convex (that follows from the convexity of the $p$-norm).

There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that $$ \int_{B_R(0)}\|u\|^pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|^pd\mathbb{P}(u). $$ Then if $x > B_{3R}(0)$, we see that $$ \int_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|^pd\mathbb{P}(u) $$ and hence is $$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u). $$ Now, just note that $x\mapsto \int_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B_{3R}(0)$. Moreover, we can say that the minimizer is unique since the function is convex (that follows from the convexity of $x\mapsto \|x\|^p$).

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hthi
hthi
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There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that $$ \int_{B_R(0)}\|u\|_pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u). $$ Then if $x > B_{3R}(0)$, we see that $$ \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u) $$ and hence is $$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u). $$ Now, just note that $x\mapsto \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B_{3R}(0)$. Moreover, we can say that therethe minimizer is a unique minimum pointunique since the function is convex (that follows from the convexity of the $p$-norm).

There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that $$ \int_{B_R(0)}\|u\|_pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u). $$ Then if $x > B_{3R}(0)$, we see that $$ \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u) $$ and hence is $$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u). $$ Now, just note that $x\mapsto \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B_{3R}(0)$. Moreover, we can say that there is a unique minimum point since the function is convex (that follows from the convexity of the $p$-norm).

There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that $$ \int_{B_R(0)}\|u\|_pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u). $$ Then if $x > B_{3R}(0)$, we see that $$ \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u) $$ and hence is $$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u). $$ Now, just note that $x\mapsto \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B_{3R}(0)$. Moreover, we can say that the minimizer is unique since the function is convex (that follows from the convexity of the $p$-norm).

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hthi
hthi
  • 415
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There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that $$ \int_{B_R(0)}\|u\|_pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u). $$ Then if $x > B_{3R}(0)$, we see that $$ \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u) $$ and hence is $$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u). $$ Now, just note that $x\mapsto \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B_{3R}(0)$. Moreover, we can say that there is a unique minimum point since the function is convex (that follows from the convexity of the $p$-norm).