$In the Cartesian coordinate plane, the minimum value of the perimeter of a convex pentagon with lattice points.
Let the 5 vertices of this convex pentagon be $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$, with coordinates $A_{j}(x_{j}, y_{j})$, and represent the vertices using complex numbers: $A_{j} = x_{j} + y_{j} i, i = 1, 2, 3, 4, 5, i$ being the imaginary unit.
Let $z_{j} = A_{j+1} - A_{j}, j = 1, 2, 3, 4, 5, A_{6} = A_{1}$, then
(1) The real and imaginary parts of $z_{j}$ are integers, and $|z_{j}| \neq 0$ (thus $|z_{j}| \geq 1$);
(2) $z_{1} + z_{2} + z_{3} + z_{4} + z_{5} = 0$;
(3) The perimeter of the convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ is $|z_{1}| + |z_{2}| + |z_{3}| + |z_{4}| + |z_{5}|$.
By convexity, any two $z_{j}, z_{k} (j \neq k)$ do not have the same direction. Also, by (1), if some $z_{j}$ has $|z_{j}| \neq 1$, then $|z_{j}|$ can only be $\sqrt{2}, 2, \sqrt{5}, \cdots$
The number of $z_{1}, z_{2}, \cdots, z_{5}$ with modulus 1 is at most 4: $\pm 1, \pm i$.
If $|z_{1}|, |z_{2}|, \cdots, |z_{5}|$ have exactly 4 ones, by (2), the remaining one would be 0, which contradicts (1).

If $|z_{1}|, |z_{2}|, \cdots, |z_{5}|$ have exactly 3 ones, and the remaining two are $\sqrt{2}$ (note that the modulus of $\sqrt{2}$ can have at most 4 values: $1 \pm i, -1 \pm i$), they would not satisfy (2). Thus, in this case, the perimeter is $\geq 1 + 1 + 1 + \sqrt{2} + 2 = 5 + \sqrt{2}$.

If $|z_{1}|, |z_{2}|, \cdots, |z_{5}|$ have exactly two ones, and the remaining three are $\sqrt{2}$, the perimeter is $1 + 1 + \sqrt{2} + \sqrt{2} + \sqrt{2} = 2 + 3\sqrt{2}$.
In other cases, the perimeter is not less than $1 + 4\sqrt{2}$.
In summary, the minimum perimeter of a lattice convex pentagon is $2 + 3\sqrt{2}$. It is not difficult to construct such a convex pentagon.