I can't find the orthogonal of this set $C= \left\{ u \in L^2(0,2): \int_0^2 u(t)dt=1 \right\}$.
By applying the standard definition I have to find some conditions on $v(x)$ such that $\int_0^2 u(x)v(x)dx=0$ for every $u \in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 \in C^{\perp}$
Any hint or else would be really appreciated
The orthogonal is trivial.
Indeed, assume $v \in C^\perp$.
If $\int_0^2 v(t)\,dt \ne 0$, then $$\int_0^2 \frac{v(x)}{\int_0^2 v(t)\,dt}\,dx = \frac{\int_0^2 v(x)\,dx}{\int_0^2 v(t)\,dt} = 1$$ so we have $\frac{v}{\int_0^2 v(t)\,dt} \in C$. Therefore $v \perp \frac{v}{\int_0^2 v(t)\,dt}$ which implies $v \perp v$ and hence $v = 0$.
If $\int_0^2 v(t)\,dt = 0$, then for any $u \in C$ we have $$\int_0^2 (u(t) + v(t))\,dt = \int_0^2 u(t)\,dt + \int_0^2 v(t)\,dt = 1$$ so $u+v \in C$. Hence $v \perp u$ and $v \perp u+v$ imply $v \perp v$. It follows $v = 0$.