Let $$f : [-1, 1] \to R$$ be defined as $$f(x) = ax^2 + bx + c$$ for all $$x \in [-1, 1]$$, where $$a, b, c \in R$$ such that $$f(-1) = 2$$, $$f'(-1) = 1$$ and for $$x \in (-1, 1)$$ the maximum value of $$f''(x)$$ is $$\frac{1}{2}$$. If $$f(x) \leq \alpha$$, $$x \in [-1, 1]$$, then the least value of $$\alpha$$ is equal to ________.

We have $$f(x) = ax^2 + bx + c$$ with $$f(-1) = a - b + c = 2$$, $$f'(x) = 2ax + b$$ so $$f'(-1) = -2a + b = 1$$, and $$f''(x) = 2a$$ with maximum value $$\frac{1}{2}$$ on $$(-1,1)$$, meaning $$2a \leq \frac{1}{2}$$, so $$a \leq \frac{1}{4}$$.

From $$f'(-1) = 1$$: $$b = 1 + 2a$$. From $$f(-1) = 2$$: $$a - (1+2a) + c = 2$$, giving $$c = 3 + a$$.

So $$f(x) = ax^2 + (1+2a)x + (3+a)$$. To find the maximum of $$f(x)$$ on $$[-1, 1]$$, we evaluate $$f(1) = a + 1 + 2a + 3 + a = 4a + 4$$.

Since $$f$$ is a quadratic with leading coefficient $$a \leq \frac{1}{4}$$, and the vertex is at $$x = -\frac{b}{2a} = -\frac{1+2a}{2a}$$, we need to check whether the maximum on $$[-1,1]$$ occurs at $$x = 1$$ or at the vertex.

For small positive $$a$$, the vertex $$x = -\frac{1+2a}{2a}$$ is very negative (far left of $$[-1,1]$$), so $$f$$ is increasing on $$[-1,1]$$ and the maximum is at $$x = 1$$. For $$a \leq 0$$, the parabola opens downward, and the vertex could be inside $$[-1,1]$$.

To maximize $$\alpha$$ (the maximum of $$f$$ on $$[-1,1]$$), we want to choose $$a$$ as large as possible since $$f(1) = 4a + 4$$ increases with $$a$$. The constraint gives $$a \leq \frac{1}{4}$$, and equality is achieved when $$f''(x) = \frac{1}{2}$$.

With $$a = \frac{1}{4}$$: $$b = 1 + \frac{1}{2} = \frac{3}{2}$$, $$c = 3 + \frac{1}{4} = \frac{13}{4}$$. The vertex is at $$x = -\frac{3/2}{1/2} = -3$$, which is outside $$[-1,1]$$, so $$f$$ is increasing on $$[-1,1]$$.

The maximum value is $$f(1) = 4 \cdot \frac{1}{4} + 4 = 5$$. Therefore the least value of $$\alpha$$ such that $$f(x) \leq \alpha$$ for all $$x \in [-1,1]$$ is $$\alpha = 5$$.