TITLE: Riesz representation theorem and integral equality in a Hilbert space
QUESTION [6 upvotes]: I'm having trouble to understand how to tackle the following problem:

Lets consider the Hilbert space $H = L^2(]0,1[)$. Show that there exist a unique $u \in L^2(]0,1[)$ such that:
$$
  \int_0^1
    \left( \int_0^xu(t) \,\mathrm{d}t \right)
    \left( \int_0^x v(t) \,\mathrm{d}t \right)
  \,\mathrm{d}x
  +
  \int_0^1 u(x)v(x) \,\mathrm{d}x
  =
  \int_0^1 f(x) v(x) \, \mathrm{d}x \,,
$$
for $f \in H$ given.

I'm unsure how to even start, I know I will use Riesz representation theorem and that I need a linear form of $H$ but this integral equality is really disturbing me. Any hints or explanation would be appreciated.
Thank you !

REPLY [8 votes]: Let
$$
B(u, v) = \int_0^1 \left(\int_0^x u\right)\left(\int_0^x v \right) + \int_0^1 uv
$$
and note that $B(u, u) \ge ||u||_{L^2}^2 \Rightarrow 
\sup_{||v|| = 1} |B(u, v)| \ge ||u||_{L^2}$, which when combined with the symmetry of $B$ implies $\sup_{||u|| = 1} |B(u, v)| > 0$. By Lax-Milgram, given any $f \in L ^2$, we will have a unique $u_f \in L ^2$ s.t. $B(u_f, v) = \langle f, v \rangle$ for all $v \in L ^2$.