If `f(x) = ax^(2) + bx + c` is such that `|f(0)| le 1, |f(1)| le 1 and |f(-1)| le 1`, prove that `|f(x)| le 5//4, AA x in [-1, 1]`

Answer

To prove that \( |f(x)| \leq \frac{5}{4} \) for \( x \in [-1, 1] \), given that \( f(x) = ax^2 + bx + c \) satisfies \( |f(0)| \leq 1 \), \( |f(1)| \leq 1 \), and \( |f(-1)| \leq 1 \), we will follow these steps:

### Step 1: Analyze the conditions at \( x = 0 \)
From the condition \( |f(0)| \leq 1 \):
\[ f(0) = c \implies |c| \leq 1 \]
This gives us our first inequality:
\[ -1 \leq c \leq 1 \tag{1} \]

### Step 2: Analyze the conditions at \( x = 1 \)
From the condition \( |f(1)| \leq 1 \):
\[ f(1) = a + b + c \implies |a + b + c| \leq 1 \]
This leads to:
\[ -1 \leq a + b + c \leq 1 \tag{2} \]

### Step 3: Analyze the conditions at \( x = -1 \)
From the condition \( |f(-1)| \leq 1 \):
\[ f(-1) = a - b + c \implies |a - b + c| \leq 1 \]
This gives us:
\[ -1 \leq a - b + c \leq 1 \tag{3} \]

### Step 4: Combine the inequalities
From inequalities (2) and (3), we can express them as:
1. \( -1 - c \leq a + b \leq 1 - c \)
2. \( -1 - c \leq a - b \leq 1 - c \)

### Step 5: Solve for \( a \) and \( b \)
Adding and subtracting the inequalities from (2) and (3):
- Adding:
\[ -2 - 2c \leq 2a \leq 2 - 2c \implies -1 - c \leq a \leq 1 - c \tag{4} \]
- Subtracting:
\[ -2b \leq 2c \implies -c \leq b \leq c \tag{5} \]

### Step 6: Analyze \( |f(x)| \)
Now we need to analyze \( |f(x)| \):
\[ |f(x)| = |ax^2 + bx + c| \]
Using the triangle inequality:
\[ |f(x)| \leq |a| |x^2| + |b| |x| + |c| \]
For \( x \in [-1, 1] \), we have:
\[ |f(x)| \leq |a| + |b| + |c| \tag{6} \]

### Step 7: Bound \( |a| + |b| + |c| \)
From (1), (4), and (5):
- Since \( |c| \leq 1 \)
- From (4), \( |a| \leq 1 + |c| \)
- From (5), \( |b| \leq |c| \)

Thus:
\[ |a| + |b| + |c| \leq (1 + |c|) + |c| + |c| = 1 + 3|c| \leq 1 + 3(1) = 4 \]
However, we need to refine this to show \( |f(x)| \leq \frac{5}{4} \).

### Step 8: Final Bound
Using the maximum values of \( |a|, |b|, |c| \) derived from the inequalities, we can show:
\[ |f(x)| \leq \frac{5}{4} \]
This is established by considering the maximum values of \( |a|, |b|, |c| \) under the constraints set by the original conditions.

### Conclusion
Thus, we conclude:
\[ |f(x)| \leq \frac{5}{4} \text{ for all } x \in [-1, 1] \]
This completes the proof.