Is the sum of two tight sequences tight? Suppose that $X_n$ and $Y_n$ are sequences of random elements defined on the same probability space $(\Omega,\mathcal F,P)$ and taking values in some separable Hilbert space $H$ (I am particularly interested in the case when $H=L^2([0,1],\mathbb R)$). We say that the sequence $X_n$ is tight if for each $\varepsilon>0$ there exists a compact subset $K_\varepsilon$ of $H$ such that
$$
P(X_n\in K_\varepsilon)\ge1-\varepsilon
$$
for each $n\ge1$.

Suppose that the sequences $X_n$ and $Y_n$ are tight. Is the sequence $X_n+Y_n$ tight?

If $X_i$ and $Y_i$ are independent for each $i\ge1$, the following proof seems to work. Let $H\times H$ denote the Cartesian product of the Hilbert space $H$ with itself. The inner-product of $H\times H$ is given by
$$
\langle (f,g),(h,k)\rangle_{H\times H}=\langle f,h\rangle_H+\langle g,k\rangle_H.
$$
Then the sequence $(X_n,Y_n)$ with values in $H\times H$ is also tight since the Cartesian product of two compact sets is a compact set and we have that
\begin{align*}
 P((X_n,Y_n)\in K_\varepsilon\times T_\varepsilon)
 &=P(X_n\in K_\varepsilon,Y_n\in T_\varepsilon)\\
 &=P(X_n\in K_\varepsilon)P(Y_n\in T_\varepsilon)\\
 &\ge (1-\varepsilon)^2.
\end{align*}
 Let us define the function $\mathcal S:H\times H\to H$ by setting $\mathcal S(f,g)=f+g$ for each $(f,g)\in H\times H$. This function is continuous since
$$
 \|f+g-(h+k)\|_H\le\|f-h\|_H+\|g-k\|_H
$$
and
$$
 \|(f,g)-(h,k)\|_{H\times H}^2=\|f-h\|_H^2+\|g-k\|_H^2.
$$
Hence, the sequence $X_n+Y_n$ is tight using the fact that if the sequence $X_n$ is tight, then so is $f(X_n)$, where $f$ is a continuous function.
Is this proof correct? Does this proof depend on the particular form of the inner product of the space $H\times H$? Is it possible to relax the assumption of independence?
Any help is much appreciated!
 A: Your proof in the case where the sequences $\left(X_n\right)_{n\geqslant 1}$ and $\left(Y_n\right)_{n\geqslant 1}$ are independent. 
In the general case, the sum $\left(X_n+Y_n\right)_{n\geqslant 1}$ is also tight. Indeed, let $\left(e_n\right)_{n\geqslant 1}$ be an orthonormal sequence. Then a sequence of random variables $\left(Z_n\right)_{n\geqslant 1}$ is tight if and only if  for each positive $\varepsilon$,
$$\lim_{J\to +\infty}\sup_{n\geqslant 1}    \mathbb P\left\{\sum_{j=J}^{+\infty}\left\langle Z_n,e_j\right\rangle^2 \geqslant \varepsilon   \right\} =0$$
and the sequence $\left(\left(\langle Z_n,e_i\rangle\right)_{i=1}^d\right)_{n\geqslant 1}$ is tight in $\mathbb R^d$ for any $d$. This is proven in 
Suquet, Ch. Tightness in Schauder decomposable Banach spaces. Proceedings of the St. Petersburg Mathematical Society, Vol. V, 201–224, Amer. Math. Soc. Transl. Ser. 2, 193, Amer. Math. Soc., Providence, RI, 1999
and which is available on the webpage of the author. The paper deals with the more general context of Schauder decomposable Banach spaces. 
