Let $\mu$ be a standard Gaussian measure on $\mathbb{R}^n$. If $f : \mathbb{R}^n \to \mathbb{R}$ belongs to the Sobolev space $H^1(\mathbb{R}^n,\mu)$, what I know is that the Gaussian Poincare inequality holds:
\begin{equation} Var(f) \leq E[ \nabla f \cdot \nabla f] \end{equation}
Now, I wonder how to generalize this inequality to the case when $f$ itself is vector-valued?
Then, the variance $Var(f)$ becomes a covariance matrix $E[ (f-E[f])(f-E[f])^T)$ but I cannot really make sense of the term $E[ \nabla f \cdot \nabla f]$. Moreover, I cannot see how to establish an inequality.
Could anyone please explain to me?
The suggestion i made in my comment indeed works:
Suppose $f:\mathbb R^d\rightarrow \mathbb R^d$. Let $x\in\mathbb R^d$ be arbitrary. Define the function $g(y):=x^Tf(y):\mathbb R^d \rightarrow \mathbb R$. Then by the gaussian Poincare inequality we have $$\text{Var}(g)\leq \mathbb E\big[(\nabla g)^T\cdot\nabla g\big]=\mathbb E\big[(\nabla f)^Txx^T\nabla f\big] = \mathbb E\big[ x^T\nabla f(\nabla f)^Tx\big]=x^T\mathbb E\big[\nabla f\cdot\nabla f^T\big]x.$$ On the other hand, note that $$\text{Var}(g)=\text{Var}(x^Tf)=\sum_i x_i^2\text{Var}(f_i)+2\sum_{i<j}x_ix_j\text{Cov}(f_i,f_j)=x^T\text{Var}(f)x$$ Since $x$ was arbitrary, this inequality holds for all $x$. which yields the relation $$\text{Var}(f)\preceq \mathbb E\big[\nabla f\cdot \nabla f^T\big]$$