Let $f: \mathbb{R^d} \rightarrow \mathbb{R}$ be a continuous function, and $\left(X_t \right)_{0\le t \le T}$ be a square integrable process with continuous path on $\left(\Omega, \mathbb{P} \right)$. Assume that $\mathbb{E}\left[\sup_{0\le t \le T}|f(X_t)|^2\right] < \infty$, then we can conclude by the Dominated Convergence Theorem that $t \mapsto f(X_t)$ is continuous from $[0,T]$ to $\mathbb{L}^2(\Omega; \mathbb{R}^d)$.
Under the above assumption can we have that $(s,t,\lambda) \mapsto f\left(X_s + \lambda(X_t - X_s) \right)$ is continuous from $[0,T]^2 \times [0,1]$ to $\mathbb{L}^2(\Omega; \mathbb{R}^d)$ ?
I tried to start from the definition of continuity. Suppose $(s_n,t_n,\lambda_n)_{n \ge 1}$ converges to $(s,t,\lambda)$ as $n \to \infty$, then it suffices to show that \begin{align*} \lim_{n \to \infty} \Vert f\left(X_{s_n} + \lambda_n(X_{t_n} - X_{s_n}) \right) - f\left(X_s + \lambda(X_t - X_s) \right) \Vert_2 = 0 \end{align*} but the above assumption seems to be not enough. Can we have a better assumption instead of assuming that $\mathbb{E}\left[\sup_{0\le t,s \le T, 0\le \lambda \le 1}|f(X_s+\lambda(X_t- X_s))|^2\right] < \infty$ ?
Please give a counterexample to show the original assumption is not sufficient if possible.
A sufficient assumption would be the uniform integrability of the collection of random variables $$ \left\{\left\lvert f\left(X_s+\lambda\left(X_t-X_s\right)\right)\right\rvert^2 ,0\leqslant s,t\leqslant T, 0\leqslant \lambda\leqslant 1\right\}. $$ Indeed, we know that a uniformly integrable sequence which converges almost everywhere converges in $\mathbb L^1$ hence using this with $$ Z_n=\left\lvert f\left(X_{s_n}+\lambda_n\left(X_{t_n}-X_{s_n}\right)\right)-f\left(X_s+\lambda\left(X_t-X_s\right)\right)\right\rvert^2 $$ gives continuity of the wanted process.