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The answer is no. E.g., let $f(x):=1+m_+$ if $|x|\in[2^{m-1},2^m)$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $m$, with $f(0):=0$. Here, $m_+:=\max(0,m)$ and $|x|$ is the Eucludean norm of $x$.

Details: Clearly, $f$ is nonnegative and nondecreasing in each of the coordinates of the argument. Take now any $x$ and $y$ in $\mathcal C$ such that $|x|\in[2^{k-1},2^k)$ and $|y|\in[2^{m-1},2^m)$ for integers $k$ and $m$ such that $k\le m$. Then $f(x)+f(y)=1+k_+ +1+m_+\ge2+m_+$ and $|x+y|\le|x|+|y|<2^k+2^m\le2^{m+1}$, whence $f(x+y)\le1+(m+1)_+\le2+m_+\le f(x)+f(y)$, so that $f$ is subadditive.

However, the condition $f(tx)\ge tf(x)$ will fail to hold if e.g. $x\in\mathcal C$, $|x|=1$, and $t\in(1/2,1)$ -- because then $f(x)=2$ whereas $f(tx)=1<tf(x)$.

The answer is no. E.g., let $f(0):=0$ and $f(x):=1+m_+$ if $|x|\in[2^{m-1},2^m)$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $m$; that is, $f(x)=1+(1+\lfloor\log_2|x|\rfloor)_+$ for all $x\in\mathcal C\setminus\{0\}$. Here, $m_+:=\max(0,m)$ and $|x|$ is the Eucludean norm of $x$.

Details: Clearly, $f$ is nonnegative and nondecreasing in each of the coordinates of the argument. Take now any $x$ and $y$ in $\mathcal C$ such that $|x|\in[2^{k-1},2^k)$ and $|y|\in[2^{m-1},2^m)$ for integers $k$ and $m$ such that $k\le m$. Then $f(x)+f(y)=1+k_+ +1+m_+\ge2+m_+$ and $|x+y|\le|x|+|y|<2^k+2^m\le2^{m+1}$, whence $f(x+y)\le1+(m+1)_+\le2+m_+\le f(x)+f(y)$, so that $f$ is subadditive.

However, the condition $f(tx)\ge tf(x)$ will fail to hold if e.g. $x\in\mathcal C$, $|x|=1$, and $t\in(1/2,1)$ -- because then $f(x)=2$ whereas $f(tx)=1<tf(x)$.

The answer is no. E.g., let $f(x):=1+n_+$ if $|x|\in[2^{n-1},2^n)$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $n$, with $f(0):=0$. Here, $n_+:=\max(0,n)$ and $|x|$ is the Eucludean norm of $x$.

Details: Clearly, $f$ is nonnegative and nondecreasing in each of the coordinates of the argument. Take now any $x$ and $y$ in $\mathcal C$ such that $|x|\in[2^{k-1},2^k)$ and $|y|\in[2^{n-1},2^n)$ for integers $k$ and $n$ such that $k\le n$. Then $f(x)+f(y)=1+k_+ +1+n_+\ge2+n_+$ and $|x+y|\le|x|+|y|<2^k+2^n\le2^{n+1}$, whence $f(x+y)\le1+(n+1)_+\le2+n_+\le f(x)+f(y)$, so that $f$ is subadditive.

However, the condition $f(tx)\ge tf(x)$ will fail to hold if e.g. $x\in\mathcal C$, $|x|=1$, and $t\in(1/2,1)$, because then $f(x)=2$ whereas $f(tx)=1<tf(x)$.

The answer is no. E.g., let $f(x):=1+m_+$ if $|x|\in[2^{m-1},2^m)$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $m$, with $f(0):=0$. Here, $m_+:=\max(0,m)$ and $|x|$ is the Eucludean norm of $x$.

Details: Clearly, $f$ is nonnegative and nondecreasing in each of the coordinates of the argument. Take now any $x$ and $y$ in $\mathcal C$ such that $|x|\in[2^{k-1},2^k)$ and $|y|\in[2^{m-1},2^m)$ for integers $k$ and $m$ such that $k\le m$. Then $f(x)+f(y)=1+k_+ +1+m_+\ge2+m_+$ and $|x+y|\le|x|+|y|<2^k+2^m\le2^{m+1}$, whence $f(x+y)\le1+(m+1)_+\le2+m_+\le f(x)+f(y)$, so that $f$ is subadditive.

However, the condition $f(tx)\ge tf(x)$ will fail to hold if e.g. $x\in\mathcal C$, $|x|=1$, and $t\in(1/2,1)$ -- because then $f(x)=2$ whereas $f(tx)=1<tf(x)$.

The answer is no. E.g., let $f(x):=1+n_+$ if $|x|\in(2^{n-1},2^n]$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $n$, with $f(0):=0$. Here, $n_+:=\max(0,n)$.

The answer is no. E.g., let $f(x):=1+n_+$ if $|x|\in[2^{n-1},2^n)$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $n$, with $f(0):=0$. Here, $n_+:=\max(0,n)$ and $|x|$ is the Eucludean norm of $x$.

Details: Clearly, $f$ is nonnegative and nondecreasing in each of the coordinates of the argument. Take now any $x$ and $y$ in $\mathcal C$ such that $|x|\in[2^{k-1},2^k)$ and $|y|\in[2^{n-1},2^n)$ for integers $k$ and $n$ such that $k\le n$. Then $f(x)+f(y)=1+k_+ +1+n_+\ge2+n_+$ and $|x+y|\le|x|+|y|<2^k+2^n\le2^{n+1}$, whence $f(x+y)\le1+(n+1)_+\le2+n_+\le f(x)+f(y)$, so that $f$ is subadditive.

However, the condition $f(tx)\ge tf(x)$ will fail to hold if e.g. $x\in\mathcal C$, $|x|=1$, and $t\in(1/2,1)$, because then $f(x)=2$ whereas $f(tx)=1<tf(x)$.