Suppose that $X$ is a random element with values in a separable Hilbert space $\mathbb H$ such that $\operatorname EX=0$ and $\operatorname E\|X\|^2<\infty$. Suppose that $f_1,f_2,\ldots$ form an orthonormal basis of $\mathbb H$ (these might be eigenvectors of the covariance operator of $X$). Then, by Parseval's identity, we have that $$ \sum_{n=1}^\infty\operatorname E|\langle X,f_n\rangle|^2<\infty. $$ In particular, $\operatorname E|\langle X,f_n\rangle|^2\to0$ as $n\to\infty$.

I am interested in $\operatorname E|\langle X,f_n\rangle|^p$ with $p>2$. If $\operatorname E\|X\|^{2(p-1)}<\infty$, using the fact that $\|f_j\|=1$ for $j\ge1$ and the Cauchy-Schwarz inequality twice, we obtain \begin{align*} \operatorname E|\langle X,f_n\rangle|^p &=\operatorname E[|\langle X,f_n\rangle||\langle X,f_n\rangle|^{p-1}]\\ &\le\operatorname E[|\langle X,f_n\rangle|\|X\|^{p-1}]\\ &\le(\operatorname E|\langle X,f_n\rangle|^2)^{1/2}(\operatorname E\|X\|^{2(p-1)})^{1/2}. \end{align*} This shows that $\operatorname E|\langle X,f_n\rangle|^p\to0$ as $n\to\infty$ as well, but at a slower rate than $\operatorname E|\langle X,f_n\rangle|^2$. The rate of convergence does not depend on $p$, which I find counterintuitive. Also, we need to assume that $\operatorname E\|X\|^{2(p-1)}<\infty$.

At the moment, I have that $\operatorname E|\langle X,f_n\rangle|^p=O((\operatorname E|\langle X,f_n\rangle|^2)^{1/2})$ as $n\to\infty$. I suspect that $\operatorname E|\langle X,f_n\rangle|^p$ goes to $0$ faster than that. Is it possible to show that $\operatorname E|\langle X,f_n\rangle|^p\to0$ as $n\to\infty$ faster than $(\operatorname E|\langle X,f_n\rangle|^2)^{1/2}$? Perhaps even $\operatorname E|\langle X,f_n\rangle|^p=O((\operatorname E|\langle X,f_n\rangle|^2)^{p/2})$ as $n\to\infty$?

Any help is much appreciated!