In Brezis 1.8 I need to show that the gauge (or Minkowski functional) of the set $$ C= \bigg\{ u \in C[0,1]: \int_{0}^{1} \vert {u(t)}^{2}\vert dt < 1 \bigg \} $$ is a norm, where $C[0,1]$ have the norm $\Vert u \Vert=\max \{ u(t) | t \in t\}$, but I don't know how to do this.
I just prove that $C$ is convex, symmetric and $0\in C$, if $C$ is open then the proof is complete, but is $C$ open? Why? I try to use that the $\int_0^1$ is continuous functional in $C[0,1]$, right?
Yes, this set is open with respect to the given norm. Given a fixed $u \in C$, we know $|u|$ is continuous on the compact interval $[0, 1]$, and hence is bounded by some $M > 0$. Define $$r = \min\left\{\frac{1}{4M + 2}\left(1 - \int_0^1 |u(s)|^2 \, \mathrm{d}s\right), 1 \right\} > 0.$$ Suppose $x \in B(u; r)$. We wish to show $x \in C$.
We have, for any $t \in [0, 1]$, \begin{align*} |x(t)|^2 &= |x(t)|^2 - |u(t)|^2 + |u(t)|^2 \\ &= (|x(t)| - |u(t)|)(|x(t)| + |u(t)|) + |u(t)|^2 \\ &= (|x(t)| - |u(t)|)(|x(t) - u(t)| + 2|u(t)|) + |u(t)|^2 \\ &\le |x(t) - u(t)|(|x(t) - u(t)| + 2|u(t)|) + |u(t)|^2 \\ &\le r(r + 2M) + |u(t)|^2 \\ &\le r(1 + 2M) + |u(t)|^2 \\ &\le \frac{1 + 2M}{4M + 2}\left(1 - \int_0^1 |u(s)|^2 \, \mathrm{d}s\right) + |u(t)|^2 \\ &= \frac{1}{2}\left(1 - \int_0^1 |u(s)|^2 \, \mathrm{d}s\right) + |u(t)|^2. \end{align*} As this holds for all $t$, we have \begin{align*} \int_0^1 |x(t)|^2 \, \mathrm{d}t &\le \frac{1}{2}\left(1 - \int_0^1 |u(s)|^2 \, \mathrm{d}s\right) + \int_0^1|u(t)|^2 \, \mathrm{d}t \\ &= \frac{1}{2}\left(1 + \int_0^1 |u(s)|^2 \, \mathrm{d}s\right) < 1. \end{align*} That is, $x \in C$, hence $B(u; r) \subseteq C$, and so $C$ is open.