Let $\mathcal{F}$ be the set of continuous functions $f : [0, 1]\to\mathbb{R}$ satisfying $\max_{0\le x\le 1} |f(x)| = 1$ and let $I : \mathcal{F} \to \mathbb{R}$,

\[I(f) = \int_0^1 f(x)\, \text{d}x - f(0) + f(1).\]

a) Show that $I(f) < 3$, for any $f \in \mathcal{F}$. 

b) Determine $\sup\{I(f) \mid f \in \mathcal{F}\}$.
### Part (a)
1. **Given Conditions:**
   - \( f \) is a continuous function on \([0, 1]\).
   - \(\max_{0 \le x \le 1} |f(x)| = 1\).

2. **Objective:**
   - Show that \( I(f) < 3 \) for any \( f \in \mathcal{F} \).

3. **Expression for \( I(f) \):**
   \[
   I(f) = \int_0^1 f(x) \, \text{d}x - f(0) + f(1)
   \]

4. **Bounding the Integral:**
   - Since \(|f(x)| \le 1\) for all \(x \in [0, 1]\), we have:
     \[
     \left| \int_0^1 f(x) \, \text{d}x \right| \le \int_0^1 |f(x)| \, \text{d}x \le \int_0^1 1 \, \text{d}x = 1
     \]
   - Therefore:
     \[
     -1 \le \int_0^1 f(x) \, \text{d}x \le 1
     \]

5. **Bounding \( f(0) \) and \( f(1) \):**
   - Since \(|f(x)| \le 1\) for all \(x \in [0, 1]\), we have:
     \[
     |f(0)| \le 1 \quad \text{and} \quad |f(1)| \le 1
     \]

6. **Combining the Bounds:**
   - We need to show that:
     \[
     I(f) = \int_0^1 f(x) \, \text{d}x - f(0) + f(1) < 3
     \]
   - Using the bounds:
     \[
     I(f) \le 1 - (-1) + 1 = 3
     \]
   - However, equality cannot hold because if \( I(f) = 3 \), then:
     \[
     \int_0^1 f(x) \, \text{d}x = 1, \quad f(0) = -1, \quad f(1) = 1
     \]
   - The condition \(\int_0^1 f(x) \, \text{d}x = 1\) implies \( f(x) \equiv 1 \) over \([0, 1]\), which contradicts \( f(0) = -1 \) since \( f \) is continuous.

7. **Conclusion:**
   - Therefore, \( I(f) < 3 \) for any \( f \in \mathcal{F} \).

\(\blacksquare\)

### Part (b)
1. **Objective:**
   - Determine \(\sup\{I(f) \mid f \in \mathcal{F}\}\).

2. **Upper Bound:**
   - From part (a), we know \( I(f) < 3 \).

3. **Constructing a Sequence:**
   - Consider the sequence of functions \( f_n(x) \) defined as:
     \[
     f_n(x) = \begin{cases} 
     1 & \text{if } x \in \left[ \frac{1}{n}, 1 \right] \\
     -1 + 2nx & \text{if } x \in \left[ 0, \frac{1}{n} \right]
     \end{cases}
     \]
   - These functions are continuous and satisfy \(\max_{0 \le x \le 1} |f_n(x)| = 1\).

4. **Calculating \( I(f_n) \):**
   - For \( f_n(x) \):
     \[
     \int_0^1 f_n(x) \, \text{d}x = \int_0^{1/n} (-1 + 2nx) \, \text{d}x + \int_{1/n}^1 1 \, \text{d}x
     \]
     \[
     = \left[ -x + nx^2 \right]_0^{1/n} + \left[ x \right]_{1/n}^1
     \]
     \[
     = \left( -\frac{1}{n} + n \left( \frac{1}{n} \right)^2 \right) + (1 - \frac{1}{n})
     \]
     \[
     = -\frac{1}{n} + \frac{1}{n} + 1 - \frac{1}{n} = 1 - \frac{1}{n}
     \]
   - Therefore:
     \[
     I(f_n) = \left( 1 - \frac{1}{n} \right) - (-1) + 1 = 3 - \frac{1}{n}
     \]

5. **Taking the Limit:**
   - As \( n \to \infty \), \( I(f_n) \to 3 \).

6. **Conclusion:**
   - Therefore, \(\sup\{I(f) \mid f \in \mathcal{F}\} = 3\).

\(\boxed{3}\)