0

Let $X$ be a Hilbert Space and let $\{S\}$ be a Convex set in $X$. Let $d=\inf_{x \in S}\|x\|$ . Prove that, if $\{x_n\}$ is a sequence of elements in $S$ such that $\lim_n \|x_n\|=d$, then $\{x_n\}$ is a Cauchy sequence.

This is a problem from the book "Functional Analysis" by Bachman and Narici ( Page-161, Q.17). I don't see why convexity of $S$ is needed and again any Normed space would have done. I guessed that all these are needed for the next question (Q.18) which is a continuation of this one.

The next Question states that :

Show that any closed convex subset $S$ of a Hilbert Space $X$ contains a unique element of the minimal form.

Suppose that $d=\inf_{x \in S}\|x\|$. Then $d+\dfrac{1}{n}$ is no longer an infimum and hence there is a $x_n \in S$ such that $\|x_n\| < d+\dfrac{1}{n}$. Then $d \le \|x_n\|=y_n < d+\dfrac{1}{n}$ and letting $n \to \infty$ we have $\lim\|x_n\|=d$

By Previous Exercise $x_n$ is a cauchy sequence in $S$ and since $S$ is closed and hence Complete implying $x_n \to x \in S$ and $\|x\|=d$.

I don't see why Convexity of $S$ is needed and a Banach Space would have done in place of Hilbert Space.

Thanks for the help!!

tattwamasi amrutam
  • 12,802
  • 5
  • 38
  • 73

0 Answers0