Convexity of $K:=\{u\in X|\int_0^1|u(t)|^2dt<1\}$ 
Hi, I have the following problem:
  I try to show the convexity of K.
  $$K:=\biggl\{u\in X|\int_0^1|u(t)|^2dt<1\biggr\}$$

Assume $a(t),b(t)\in K$. Now I tried to show the definition:
$$\forall 0\le\lambda\le1: \lambda a(t)+(1-\lambda) b(t)\in K.$$
$$\int_0^1|u(t)|^2dt$$$$\int_0^1|\lambda a(t)+(1-\lambda)b(t)|^2dt$$$$=\int_0^1|(\lambda a(t))^2+2(-\lambda^2+\lambda) a(t)b(t)+(\lambda b(t))^2-2\lambda b(t)^2+b(t)^2|dt$$ Now I have problems to continue. Could someone help me here?
 A: Since $x \mapsto x^2$ is convex, if $\lambda \in [0,1]$ then
$(\lambda u_1(t)+(1-\lambda) u_2(t))^2 \le \lambda u_1(t)^2+(1-\lambda) u_2(t)^2$.
Now integrate over $[0,1]$.
A: Alternative approach: the polynomials $\frac{P_n(2x-1)}{\sqrt{2n+1}}$ provide an orthonormal base of $L^2(0,1)$ with respect to the standard inner product. Assuming 
$$ f(x)\stackrel{L^2}{=}\sum_{n\geq 0} f_n\cdot\frac{P_n(2x-1)}{\sqrt{2n+1}},\qquad g(x)\stackrel{L^2}{=}\sum_{n\geq 0} g_n\cdot\frac{P_n(2x-1)}{\sqrt{2n+1}} $$
we have $f\in K$ iff $\sum_{n\geq 0}f_n^2 < 1$ (this is just the usual isometry between $L^2$ and $\ell^2$).
If both $f$ and $g$ belong to $K$, for any $\lambda\in[0,1]$ we have
$$ \|\lambda f+(1-\lambda)g\|_{2}^{2} = \sum_{n\geq 0}\left(\lambda f_n+(1-\lambda) g_n\right)^2\leq \lambda\|f\|_2^2+(1-\lambda)\|g\|_2^2<1 $$
by the convexity of $x\mapsto x^2$, ensuring $(\lambda u+(1-\lambda) v)^2\leq \lambda u^2+(1-\lambda)v^2$.
It follows that $K$ is convex as claimed. The unit ball is convex in any Hilbert space equipped with a positive-definite inner product, for the very same reason.
