inequality with sup and Expectations

Question

let $X_{t}$ Itô diffusion how we can get :

$$ \sup_{x\in \mathbb{R}^{n}}|E^{x}[f(X_{t})]| \leq \sup_{y \in \mathbb{R}^{n}}|f(y)|\sup_{x\in\mathbb{R}}E^{x}[1]=\sup_{y\in\mathbb{R}^{n}}|f(y)|.$$

Here $E^{x}$ denotes the expectation w.r.t. the probability measure $P^{x}$ Thus $E[f(X_{h})]$ means $E[f(X_{h}^{y})]$, where $E$ denotes the expectation w.r.t. the measure $P$.

Answer 1

Let $\|f\|_\infty=\sup\{|f(y)|\,;\,y\in\mathbb R^n\}$ and fix some $t$. Then $|f(X^x_t(\omega))|\leqslant\|f\|_\infty$ for every $x$ and $\omega$ hence $E[|f(X^x_t)|]\leqslant\|f\|_\infty$ for every $x$. Since $|E[f(X^x_t)]|\leqslant E[|f(X^x_t)|]$ for every $x$, this implies $|E[f(X^x_t)]|\leqslant\|f\|_\infty$ for every $x$. That is, $\sup\{|E[f(X^x_t)]|\,;\,x\in\mathbb R^n\}\leqslant\|f\|_\infty$.