The answer to the question is No. Let $x:[0,1] \to [0,1]$ be defined by $x(t) = 0$ for $t\in [0,1/2[$ and $x(t) = 1$ for $t\in[1/2, 1]$. Clearly, $x$ is cadlag and therefore a member of $D([0,1], R)$.

Note that for all $\epsilon > 0$ we have $f_\epsilon(t) < t$ for $t\in]0,1[$.

So for all $\epsilon >0$ we have $(x\circ f_\epsilon)(1/2) = 0$ and $x(1/2) = 1$ therefore $1 \leq \lim_{\epsilon\to 0} ||x\circ f_\epsilon -x|| \leq \lim_{\epsilon\to 0} ||x\circ f_\epsilon -x|| + ||x\circ b_\epsilon -x ||$.

The answer to the question is No. Let $x:[0,1] \to [0,1]$ be defined by $x(t) = 0$ for $t\in [0,1/2[$ and $x(t) = 1$ for $t\in[1/2, 1]$. Clearly, $x$ is cadlag and therefore a member of $D([0,1], R)$.

Note that for all $\varepsilon > 0$ we have $f_\varepsilon(t) < t$ for $t\in]0,1[$.

So for all $\epsilon >0$ we have $(x\circ f_\epsilon)(1/2) = 0$ and $x(1/2) = 1$ therefore $1 \leq \lim_{\varepsilon\to 0} ||x\circ f_\varepsilon -x|| \leq \lim_{\varepsilon\to 0} ||x\circ f_\varepsilon -x|| + ||x\circ b_\varepsilon -x ||$.

The answer to the question is No. Let $x:[0,1] \to [0,1]$ be defined by $x(t) = 0$ for $t\in [0,1/2[$ and $x(t) = 1$ for $t\in[1/2, 1]$. Clearly, $x$ is cadlag and therefore a member of $D([0,1], R)$.

Note that for all $\epsilon > 0$ we have $f_\epsilon(t) < t$ for $t\in]0,1[$.

So for all $\epsilon >0$ we have $(x\circ f_\epsilon)(1/2) = 0$ and $x(1/2) = 1$ therefore $1 \leq \lim_{\epsilon\to 0} ||x\circ f_\epsilon || \leq \lim_{\epsilon\to 0} ||x\circ f_\epsilon || + ||x\circ b_\epsilon ||$.

The answer to the question is No. Let $x:[0,1] \to [0,1]$ be defined by $x(t) = 0$ for $t\in [0,1/2[$ and $x(t) = 1$ for $t\in[1/2, 1]$. Clearly, $x$ is cadlag and therefore a member of $D([0,1], R)$.

Note that for all $\epsilon > 0$ we have $f_\epsilon(t) < t$ for $t\in]0,1[$.

So for all $\epsilon >0$ we have $(x\circ f_\epsilon)(1/2) = 0$ and $x(1/2) = 1$ therefore $1 \leq \lim_{\epsilon\to 0} ||x\circ f_\epsilon -x|| \leq \lim_{\epsilon\to 0} ||x\circ f_\epsilon -x|| + ||x\circ b_\epsilon -x ||$.