Given two points $ P,\ Q$ on the parabola $ y \equal{} x^2$. Let $ h$ be the $ y$ coordinate of the midpoint of the segment $ PQ$.

(1) Let $ L$ and $ m$ be the length and slope of the segment $ PQ$ respectively. Express $ h$ in terms of $ L,\ m$.

(2) When $ L$ is fixed, find the minimum value of $ h$.

[color=green]Corrected by moderator.[/color]
Given two points \( P(p, p^2) \) and \( Q(q, q^2) \) on the parabola \( y = x^2 \), we need to express \( h \) in terms of \( L \) and \( m \), and then find the minimum value of \( h \) when \( L \) is fixed.

### Part 1: Express \( h \) in terms of \( L \) and \( m \)

1. **Slope \( m \) of the segment \( PQ \)**:
   \[
   m = \frac{q^2 - p^2}{q - p} = p + q
   \]

2. **Midpoint \( M \) of the segment \( PQ \)**:
   \[
   M = \left( \frac{p+q}{2}, \frac{p^2 + q^2}{2} \right)
   \]
   Therefore, the \( y \)-coordinate of the midpoint \( M \) is:
   \[
   h = \frac{p^2 + q^2}{2}
   \]

3. **Length \( L \) of the segment \( PQ \)**:
   \[
   L = \sqrt{(q - p)^2 + (q^2 - p^2)^2} = \sqrt{(q - p)^2 \left[ 1 + (p + q)^2 \right]}
   \]
   Squaring both sides, we get:
   \[
   L^2 = (q - p)^2 \left[ 1 + (p + q)^2 \right]
   \]

4. **Express \( (q - p)^2 \) in terms of \( L \) and \( m \)**:
   \[
   (q - p)^2 = \frac{L^2}{1 + m^2}
   \]

5. **Express \( p^2 + q^2 \) in terms of \( L \) and \( m \)**:
   \[
   p^2 + q^2 = (p + q)^2 - 2pq = m^2 - 2pq
   \]
   Since \( (q - p)^2 = q^2 - 2pq + p^2 \), we have:
   \[
   q^2 + p^2 = (q - p)^2 + 2pq = \frac{L^2}{1 + m^2} + 2pq
   \]
   But we need \( p^2 + q^2 \) directly, so:
   \[
   p^2 + q^2 = \frac{L^2}{1 + m^2} + m^2
   \]

6. **Express \( h \) in terms of \( L \) and \( m \)**:
   \[
   h = \frac{p^2 + q^2}{2} = \frac{1}{2} \left( \frac{L^2}{1 + m^2} + m^2 \right)
   \]
   Therefore:
   \[
   4h = \frac{L^2}{1 + m^2} + m^2
   \]

### Part 2: Find the minimum value of \( h \) when \( L \) is fixed

1. **Rewrite the expression for \( h \)**:
   \[
   4h = \frac{L^2}{1 + m^2} + m^2
   \]

2. **Analyze the expression for \( h \)**:
   \[
   4h = \frac{L^2}{1 + m^2} + m^2
   \]
   To minimize \( h \), we need to minimize the right-hand side of the equation.

3. **Consider the function \( f(m) = \frac{L^2}{1 + m^2} + m^2 \)**:
   \[
   f'(m) = -\frac{2L^2m}{(1 + m^2)^2} + 2m
   \]
   Setting \( f'(m) = 0 \):
   \[
   -\frac{2L^2m}{(1 + m^2)^2} + 2m = 0
   \]
   \[
   -\frac{L^2m}{(1 + m^2)^2} + m = 0
   \]
   \[
   m \left( 1 - \frac{L^2}{(1 + m^2)^2} \right) = 0
   \]
   Therefore, \( m = 0 \) or \( 1 = \frac{L^2}{(1 + m^2)^2} \).

4. **Case \( m = 0 \)**:
   \[
   4h = \frac{L^2}{1 + 0^2} + 0^2 = L^2
   \]
   \[
   h = \frac{L^2}{4}
   \]

5. **Case \( 1 = \frac{L^2}{(1 + m^2)^2} \)**:
   \[
   (1 + m^2)^2 = L^2
   \]
   \[
   1 + m^2 = L
   \]
   \[
   m^2 = L - 1
   \]
   \[
   4h = \frac{L^2}{L} + (L - 1) = L + L - 1 = 2L - 1
   \]
   \[
   h = \frac{2L - 1}{4}
   \]

6. **Compare the two cases**:
   \[
   \text{For } L < 1, \quad h = \frac{L^2}{4}
   \]
   \[
   \text{For } L \geq 1, \quad h = \frac{2L - 1}{4}
   \]

The final answer is \( \boxed{h = \frac{L^2}{4} \text{ for } L < 1 \text{ and } h = \frac{2L - 1}{4} \text{ for } L \geq 1} \)