Proving hypotheses in order to deduce the Minkowski functional of $C = := \{ u \in X: \int_{0}^{1} |u(x)|^2 dx < 1\}$ I attempted to solve this problem.
Let $X = C^0([0,1])$ endowed with the norm $||\cdot||_{C^0}$. Let $C := \{ u \in X: \int_{0}^{1} |u(x)|^2 dx < 1\}$.

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*Prove that $C$ is open, convex and that $0 \in C$.

*Discuss whether $C$ is bounded or not.

I proceeded as follows.
Proving that $0 \in C$ is immediate since, if $u(x) = 0, \forall x \in [0,1]$, then $\int_{0}^{1} |u(x)|^2 dx = 0 < 1$.
In order to demonstrate that $C$ is convex, I observed that $\int_{0}^{1} |u(x)|^2 dx =: ||u||^2_{L^2([0,1])} < 1$ and, hence, $||u||_{L^2([0,1])} < 1$.
By definition, $C$ is convex if $\forall u,v \in C$ and $\forall \lambda \in [0,1]$, $\lambda u + (1-\lambda) v \in C$. Thus,
$\int_{0}^{1} |\lambda u(x) + (1-\lambda) v(x)|^2 dx \le \int_{0}^{1} (\lambda |u(x)| + (1-\lambda) |v(x)|)^2 dx \le \lambda^2 \int_{0}^{1} |u(x)|^2 dx + (1-\lambda)^2 \int_{0}^{1} |v(x)|^2 dx + 2 \lambda \, (1-\lambda) \int_{0}^{1} |u(x)| \cdot |v(x)| dx < \lambda^2 + 1 + \lambda^2 -2\lambda + 2\lambda(1-\lambda) < 1$, where $\int_{0}^{1} |u(x)| \cdot |v(x)| dx = \int_{0}^{1} |u(x) \cdot v(x)| dx =  ||u-v||_{L^1([0,1])}$ and, according to Holder's inequality, $||u-v||_{L^1([0,1])} \le ||u||_{L^2([0,1])} ||v||_{L^2([0,1])} < 1$.
However, I am struggling proving that $C$ is open. I know I have to show that $\forall u \in C, \exists r > 0: B_{r}(u) \subset C$ which means that, for any $v \in B_r(u)$, that is $||v - u||_{C^0} < r$, $v \in C$. My idea initially was to estimate $||\cdot||_{C^0}$ with the $L^2-$norm, but I actually showed that the two norms are not topologically equivalent and, therefore, I have no clue on how to proceed.
Similarly, I got stuck in establishing whether $C$ is bounded or not ($C$ is bounded $\Leftrightarrow$ $\exists r > 0: \forall u, v \in C, ||u-v||_{C^0} < r$).
Any suggestion or hint on how to conclude would be really appreciated.
 A: Suppose $||v-u||_{C^0} < r$. Define $w=v-u$.
Then we get $||w||_2^2 =\int_0^1 |w|^2< \int r^2 = r^2$.
Then by triangular equation you get
$$ ||v||_2\leq ||u||_2+||w||_2 < ||u||_2+r$$
Now if $||u||_2<1$ you can easily find an $r$ to that the right side is still smaller than $1$.
Meanwhile the set is not bounded. You can easily construct an $u$ with arbitrarily high supremum, but arbitrarily low $2$-norm. For example $u_n(x) = \max(0,n-n^3x)$. Then $||u_n||_{C^0}=n$, but
$$ \int_0^1 u_n^2 = \int_0^{1/n^2} (n-n^3x)^2 <\int_0^{1/n^2} n^2 = n^2/n^2 = 1 $$
A: In order to prove that $C$ is $\textbf{not}$ bounded, consider the functions $u_n(x) := \frac{\sqrt{n}}{1+nx}$, defined for all $n \in \mathbb{N}$.
Observe that $u_n \in C$ since $\int_{0}^{1} |u_n(x)|^2 dx = \int_{0}^{1} \frac{n}{(1+nx)^2} dx <1, \forall n \in \mathbb{N}$. However, we have that $||u_n||_{C^0} := \max_{x \in [0,1]} |u_n(x)| = \sqrt{n}$, which diverges for $n \rightarrow \infty$.
This solution can be found in Brèzis, $\textit{Functional analysis, Sobolev Spaces and Partial differential equations}$, Ex. 1.8.
