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Let $K\subset\mathbb{R}^n$ be a closed convex set. Let $p\in (1,\infty)$, $\delta\geq 0$ and $F_\delta:\mathbb{R}^n\rightarrow\mathbb{R}$ defined by $$F_\delta(x)=(\delta^2+|x|^2)^{\frac{p}{2}}$$

Note that $F_\delta$ is strictly convex, hence, we can find for all fixed $\delta$, a unique point $x_\delta\in K$ such that $$\inf_{x\in K}F_\delta(x)=F(x_\delta)$$

Denote $x_0=x$ and $F_0=F$. How can one show that $x_\delta\rightarrow x$.

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up vote 1 down vote accepted

Since the level sets of $F_\delta$ do not depend on $\delta$ or on $p$, it seems that $x_\delta = x$ for all $\delta$. In words, $x_\delta$ is the point on $K$ that is closest to the origin.

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Please, can you clarify your answer? – Tomás Nov 29 '12 at 13:22
Fix $\delta$ and $p$ and define $g(t) = \sqrt{t^{2/p} - \delta^2}$ for $t > \delta^p$. This is a monotone function of $t$. The function $x \mapsto F_\delta(x)$ attains its minimum on $K$ therefore at the same point as the function $x \mapsto g(F_\delta(x))$. But $g(F_\delta(x)) = |x|$. – Hans Engler Nov 29 '12 at 13:41

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