Relation between boundedness, Lipschitz continuity and linear growth condition of a function

Given a function $f(\cdot):[a, b]\times \mathbb{R}^n\times \mathbb{R}^m\rightarrow \mathbb{R}^n$ which is continuous on its domain.

(1) If $f(\cdot)$ is bounded on its domain, then can we say it satisfies linearity growth condition with respect to second and third arguments of the type

$\hspace{1 cm}\|f(t, x, u)\|\leq c_1\|x\|+c_2\|u\|+c_3$ for some non-negative constants $c_1, c_2, c_3?$

(2) If $f(\cdot)$ satisfies Lipschitz continuity with respect to second and third arguments i.e

$\hspace{1 cm} \|f(t, x_1, u_1)-f(t, x_2, u_2)\|\leq l_1\|x_1-x_2\|+l_2\|u_1-u_2\|,$

then can we say that $f(\cdot)$ satisfies linearity growth condition with respect to second and third arguments as mentioned in (1)?

I found the answer for this question. But I want the clarification for this answer. Can somebody tell me the following procedure is correct?

(1) Since $f(\cdot)$ is bounded on its domain, so there exists a non-negative real number $K$ such that

$\hspace{0.5 cm} \|f(t, x, u)\|\leq K.$

Thus we can write $\|f(t, x, u)\|\leq 0\|x\|+0\|u\|+K$ which confirms that $f(\cdot)$ satisfies the growth condition with respect to second and third arguments.

(2) Since $f(\cdot)$ satisfies the Lipschitz condition with respect to second and third arguments, so in particular we have

$\hspace{0.5 cm} \|f(t, x, u)-f(t, \mathbf{0}, \mathbf{0})\|\leq l_1\|x\|+l_2\|u\|$

which can be written as

$\hspace{0.5 cm}\|f(t, x, u)\|\leq \|f(t, \mathbf{0}, \mathbf{0})\|+l_1\|x\|+l_2\|u\|.$

Since $f(\cdot)$ is continuous on its domain, so in particular $f(\cdot):[a, b]\times \{\mathbf{0}\}\times \{\mathbf{0}\}\rightarrow \mathbb{R}^n$ must also be continuous and hence bounded on $[a, b]\times \{\mathbf{0}\}\times \{\mathbf{0}\},$ so $\|f(t, \mathbf{0}, \mathbf{0})\|\leq L$ and hence the above inequality becomes

$\hspace{0.5 cm}\|f(t, x, u)\|\leq L+l_1\|x\|+l_2\|u\|.$

This confirms again $f(\cdot)$ satisfies the growth condition with respect to second and third arguments.