Bounds of a Twice Differentiable Function Satisfying Differential Inequality

To determine the bounds of $f(x)$ for $0 < x < 1$, we construct an auxiliary function and analyze its convexity using the given differential inequality.

Let us define the function $g : [0, 1] \rightarrow \mathbb{R}$ as: $g(x) = e^{-x} f(x)$

Differentiating $g(x)$ with respect to $x$: $g'(x) = e^{-x} f'(x) - e^{-x} f(x) = e^{-x} \big( f'(x) - f(x) \big)$

Differentiating a second time: $g''(x) = e^{-x} \big( f''(x) - f'(x) \big) - e^{-x} \big( f'(x) - f(x) \big)$ $g''(x) = e^{-x} \big( f''(x) - 2f'(x) + f(x) \big)$

We are given the differential inequality: $f''(x) - 2f'(x) + f(x) \ge e^x, \quad \forall x \in [0, 1]$

Substituting this into the expression for $g''(x)$: $g''(x) \ge e^{-x} \cdot e^x = 1 > 0, \quad \forall x \in [0, 1]$

Since $g''(x) > 0$ for all $x \in [0, 1]$, $g(x)$ is a strictly convex function on the interval $[0, 1]$.

Now, evaluating $g(x)$ at the boundary points $x = 0$ and $x = 1$: $g(0) = e^0 \cdot f(0) = 1 \cdot 0 = 0$ $g(1) = e^{-1} \cdot f(1) = e^{-1} \cdot 0 = 0$

By the definition of strict convexity, the graph of $g(x)$ on the open interval $(0, 1)$ lies strictly below the secant line connecting its endpoints $(0, g(0))$ and $(1, g(1))$.

Since $g(0) = g(1) = 0$, the secant line is $y = 0$. Therefore: $g(x) < 0 \quad \text{for all } x \in (0, 1)$

Recalling $g(x) = e^{-x} f(x)$: $e^{-x} f(x) < 0 \quad \text{for all } x \in (0, 1)$

Since $e^{-x} > 0$ for all real $x$, we can divide by $e^{-x}$ without changing the direction of the inequality: $f(x) < 0 \quad \text{for all } x \in (0, 1)$

Thus, for $0 < x < 1$, the function satisfies: $-\infty < f(x) < 0$

Correct Option: D