Is $K:=\{u\in X|\int_0^1|u(t)|^2dt<1\}$ bounded?

Question

Let $X:=C([0,1];\mathbb{R})$ with the norm $||u||_\infty:=sup_{t\in[0,1]}|u(t)|$. Also let $K:=\{u\in X|\int_0^1|u(t)|^2dt<1\}$

Show that K is convex, symmetric, $0\in K$. And show if K is bounded.

I already showed the convexity and that K is symmetric and contains $0$. But I have no idea how I would find a boundary for K. Can someone help me?

If $u \in C([0,1])$ with $|u |_\infty > 1$, would we have $u \in K$? — Demophilus, Nov 08 '17 at 21:57

score 2 · Accepted Answer · answered Nov 08 '17 at 23:02

2

Consider the familly of functions defined by

$$\left\{ \begin{array} .f_n(x) = \frac{n-n^2x}{2} & 0 \leq x \leq \frac{1}{n^2} \\ f_n(x) = 0 & \frac{1}{n^2} \leq x \end{array} \right.$$

$\| f_n \|_\infty = \frac{n}{2} \to +\infty$ but $f_n \in K$ :

$$\int_0^1 |f_n(x)|^2 dx = 1 + \frac{1}{3 n^2} - \frac{1}{n} < 1$$

answered Nov 08 '17 at 23:02

Tryss

14,310

Sorry, I misread the question before. I thought we were talking about the $L^2$ norm, but we are not. Apologies. – Jack D'Aurizio Nov 09 '17 at 00:54

Is $K:=\{u\in X|\int_0^1|u(t)|^2dt<1\}$ bounded?

1 Answers1