Convergence in Distribution of a random function Let $H$ be a hilbert space and suppose that we have a sequence of random elements $\chi_n\in H$, $\chi_0\in H$. Let $\{\phi_j\}_{j\in\mathbb{N}}$ be an complete orthonormal basis in $H$. Suppose that I know that $\int \chi_n \phi_j \xrightarrow{D} \int \chi_0\phi_j$ for each fixed $j$. 
Question: Does this tell me that $\chi_n$ converges weakly to $\chi_0$?
 A: Edited Counterexample
I am still not sure what you mean by $\xrightarrow{D}$ but if I interpret it as a limit in $\mathbb{R}$ or $\mathbb{C}$ then I have an adequate counterexample now.
Notation: $(f,g) = \int f \cdot\bar{g} \,\mathrm{d}\lambda$ (the Hilbertspace scalarproduct)
Assumption: There is a sequence $\chi_n$ in $H$ and a vector $\chi_0 \in H$ such that $$\lim_{n\in \mathbb{N}}(\chi_n - \chi_0,\phi_i)=0\quad\text{for each fixed}\quad j \in \mathbb{N}$$
Where $\phi_i$ is an ONB of $H$.
Now choose $\chi_n = n\cdot \phi_n$, $\;\chi_0 = 0$ and $g = \sum_{j=1}^\infty \frac{1}{j}\phi_j$. Then the condtion
$$(\chi_n,\phi_i) \to (\chi_0,\phi_i)$$
is satisfied. Clearly $(\chi_0,g) = 0$ but 
$$\lim_{n\in\mathbb{N}}(\chi_n,g) = \lim_{n\in\mathbb{N}}\sum_{j=1}^\infty\frac{1}{i}(n\phi_n,\phi_i) = \lim_{n\in\mathbb{N}}\frac{1}{n}n (\phi_n,\phi_n) = 1 \neq 0 = (\chi_0,g)$$
So there is a $g \in H$ such that $\lim_{n\in\mathbb{N}}(\chi_n,g) \neq (\chi_0,g)$ which contradicts the weak convergence.
Old Post:
As I finished my counterexample I noticed that I ignored that there is an element $\chi_0 \in H$. In my counterexample there can't be such a $\chi_0$. But maybe you can see what this $\chi_0$ must give you.
Counterexample
Choose $\chi_n = \sum_{j=1}^n \phi_j$ and $g = \sum_{j=1}^\infty \frac{1}{j}\phi_j$. then you know $(\chi_n,\phi_j) \to 1$ but $$(\chi_n,g) = \sum_{j=1}^n \frac{1}{j} $$
which will not converge. So $\chi_n$ doesn't converge weakly.
