Return to Answer

The answer is no to both questions. E.g., suppose that $T=2$ and $$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$ Then $f_n(t)$ increases to $f(t)=1(t<1)$, and all the other conditions on $f_n$ hold.

However, by the definition of the Skorokhod metric, for each integer $n\ge1$, $t_n:=1-\tfrac1{2n}$, and some strictly increasing continuous function $h_n\colon[0,2]\to[0,2]$ we have
$$d(f_n,f)+\tfrac1n\ge|f_n(t_n)-f(h_n(t_n))|=|\tfrac12-f(h(t_n))| =\tfrac12,$$ since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f_n,f)\not\to0$.

The answer is no to both questions. E.g., suppose that $T=2$ and $$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$ Then $f_n(t)$ increases in $n$ to $f(t)=1(t<1)$, and all the other conditions on $f_n$ hold.

However, by the definition of the Skorokhod metric, for each integer $n\ge1$, $t_n:=1-\tfrac1{2n}$, and some strictly increasing continuous function $h_n\colon[0,2]\to[0,2]$ we have
$$d(f_n,f)+\tfrac1n\ge|f_n(t_n)-f(h_n(t_n))|=|\tfrac12-f(h(t_n))| =\tfrac12,$$ since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f_n,f)\not\to0$.

The idea of this example is simple: the range of each continuous decreasing function $f$ is connected, but the range of the right-continuous pointwise limit $f$ does not have to be connected, even if $f_n$ increases to $f$.

It is easy to modify this example to make $f_n(t)$ strictly decreasing in $n$ and in $t$, if so desired.

The answer is no to both questions. E.g., suppose that $T=2$ and $$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$ Then $f_n(t)$ increases to $f(t)=1(t<1)$, and all the other conditions on $f_n$ hold.

However, by the definition of the Skorokhod metric, for $t_n:=1-\tfrac1{2n}$ and any strictly increasing continuous function $h\colon[0,2]\to[0,2]$ we have
$$d(f_n,f)\ge|f_n(t_n)-f(h(t_n))|=|\tfrac12-f(h(t_n))| =\tfrac12,$$ since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f_n,f)\not\to0$.

The answer is no to both questions. E.g., suppose that $T=2$ and $$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$ Then $f_n(t)$ increases to $f(t)=1(t<1)$, and all the other conditions on $f_n$ hold.

However, by the definition of the Skorokhod metric, for each integer $n\ge1$, $t_n:=1-\tfrac1{2n}$, and some strictly increasing continuous function $h_n\colon[0,2]\to[0,2]$ we have
$$d(f_n,f)+\tfrac1n\ge|f_n(t_n)-f(h_n(t_n))|=|\tfrac12-f(h(t_n))| =\tfrac12,$$ since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f_n,f)\not\to0$.

The answer is no to both questions. E.g., suppose that $T=2$ and $$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$ Then $f_n(t)$ increases to $f(t)=1(t<0)$, and all the other conditions on $f_n$ hold.

However, by the definition of the Skorokhod metric, for $t_n:=1-\tfrac1{2n}$ and any strictly increasing continuous function $h\colon[0,2]\to[0,2]$ we have
$$d(f_n,f)\ge|f_n(t_n)-f(h(t_n))|=|\tfrac12-f(h(t_n))| \ge\tfrac12,$$ since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f_n,f)\not\to0$.

The answer is no to both questions. E.g., suppose that $T=2$ and $$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$ Then $f_n(t)$ increases to $f(t)=1(t<1)$, and all the other conditions on $f_n$ hold.

However, by the definition of the Skorokhod metric, for $t_n:=1-\tfrac1{2n}$ and any strictly increasing continuous function $h\colon[0,2]\to[0,2]$ we have
$$d(f_n,f)\ge|f_n(t_n)-f(h(t_n))|=|\tfrac12-f(h(t_n))| =\tfrac12,$$ since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f_n,f)\not\to0$.