Proving The Integral Inequality ∫[0 To 2] (f''(x))^2 Dx ≥ 3/(a^2-3a+4)

The fascinating world of mathematical analysis often presents us with intricate challenges that require a blend of techniques and ingenuity to unravel. One such challenge lies in the realm of integral inequalities, where we seek to establish relationships between integrals and other mathematical expressions. In this article, we delve into a specific integral inequality problem that involves a twice-differentiable function, boundary conditions, and a quest to prove a lower bound for the integral of the square of its second derivative. This exploration will not only showcase the power of tools like the Cauchy-Schwarz inequality but also illuminate the beauty of mathematical reasoning in connecting seemingly disparate concepts.

This article will explore a specific problem in integral inequalities. Our focus is on a twice-differentiable function f defined on the interval [0, 2], subjected to specific conditions. The heart of the matter lies in proving a lower bound for the integral of the square of its second derivative, specifically: ∫[0 to 2] (f''(x))^2 dx ≥ 3/(a^2-3a+4). The given conditions add layers of complexity and richness to the problem. We are told that f(0) - (a + 1)f(1) + af*(2) = 1 and that f'(2) = 0, where a is a positive real number. These boundary conditions act as constraints, shaping the behavior of the function f and influencing the possible values of the integral we aim to bound. The inequality we aim to prove is not merely an abstract mathematical statement; it has potential implications in various fields, such as physics and engineering, where the behavior of functions and their derivatives often plays a crucial role. Understanding such inequalities can provide insights into the stability, regularity, and qualitative properties of solutions to differential equations and other mathematical models.

Before diving into the solution, let's formally state the problem we intend to solve. This will provide a clear roadmap for our mathematical journey.

Let $f$ be a twice differentiable function on $[0,2]$ such that:

$$f(0)-(a+1)f(1)+af(2)=1, \quad f'(2)=0\quad\forall a >0.$$

Prove that:

$$\int_0^2 [f''(x)]^2 dx \geq \frac{3}{a^2-3a+4}.$$

The problem presents us with a twice-differentiable function f defined on the interval [0, 2]. The function f is not just any function; it is constrained by two specific conditions. The first condition, f(0) - (a + 1)f(1) + af*(2) = 1, is a linear combination of the function's values at three distinct points: 0, 1, and 2. This condition can be viewed as a weighted average of the function's values, with the weights depending on the parameter a. The fact that this weighted average is equal to 1 imposes a global constraint on the function's behavior across the entire interval [0, 2]. The second condition, f'(2) = 0, provides information about the function's derivative at the endpoint 2. It states that the slope of the tangent line to the graph of f at x = 2 is zero. This condition suggests that the function may have a local extremum (maximum or minimum) or an inflection point at x = 2. The heart of the problem lies in proving the inequality ∫[0 to 2] (f''(x))^2 dx ≥ 3/(a^2-3a+4). This inequality provides a lower bound for the integral of the square of the second derivative of f. The second derivative, f''(x), represents the concavity of the function f. The integral of its square over the interval [0, 2] can be interpreted as a measure of the overall "wiggliness" or curvature of the function. The inequality states that this wiggliness is bounded below by a specific expression involving the parameter a. This lower bound provides valuable information about the function's behavior and its relationship to the parameter a.

To tackle this problem, we'll employ a clever strategy that combines integration by parts, the Cauchy-Schwarz inequality, and a bit of algebraic manipulation. Here's the roadmap:

1. Strategic Integration: We'll start by strategically integrating expressions involving f''(x) to relate them to f'(x) and f(x).
2. Harnessing Cauchy-Schwarz: The Cauchy-Schwarz inequality will be our key tool for establishing a lower bound for the integral.
3. Algebraic Manipulation: We'll carefully manipulate the resulting expressions to arrive at the desired inequality.

Step 1: Strategic Integration

Let's define a function $g(x)$ as follows:

$$g(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2-x, & 1 \leq x \leq 2 \end{cases}$$

Now, we'll compute the integral of $f''(x)g(x)$ over the interval [0, 2] using integration by parts. The choice of g(x) is crucial here. This piecewise linear function is designed to interact favorably with the given boundary condition f(0) - (a + 1)f(1) + af*(2) = 1. By carefully choosing g(x), we set the stage for exploiting this condition later in the solution.

$$\int_0^2 f''(x)g(x) dx = \int_0^1 f''(x)x dx + \int_1^2 f''(x)(2-x) dx$$

Applying integration by parts to each integral:

$$\int_0^1 f''(x)x dx = [f'(x)x]_0^1 - \int_0^1 f'(x) dx = f'(1) - [f(x)]_0^1 = f'(1) - f(1) + f(0)$$

$$\int_1^2 f''(x)(2-x) dx = [f'(x)(2-x)]_1^2 + \int_1^2 f'(x) dx = -f'(1) + [f(x)]_1^2 = -f'(1) + f(2) - f(1)$$

Adding these two results:

$$\int_0^2 f''(x)g(x) dx = f'(1) - f(1) + f(0) - f'(1) + f(2) - f(1) = f(0) - 2f(1) + f(2)$$

Notice that the expression f(0) - 2f(1) + f(2) appears in the result. This is a key observation, as it closely resembles the given condition f(0) - (a + 1)f(1) + af*(2) = 1. We are getting closer to leveraging the given information.

Step 2: Deft Application of Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is a powerful tool for establishing relationships between integrals. Cauchy-Schwarz inequality will allow us to relate the integral of the product of two functions to the integrals of their squares. It states that for any two functions u(x) and v(x):

$$\left( \int_a^b u(x)v(x) dx \right)^2 \leq \left( \int_a^b [u(x)]^2 dx \right) \left( \int_a^b [v(x)]^2 dx \right)$$

In our case, let's choose u(x) = f''(x) and v(x) = g(x). Applying the Cauchy-Schwarz inequality to the integral we computed in the previous step:

$$\left( \int_0^2 f''(x)g(x) dx \right)^2 \leq \left( \int_0^2 [f''(x)]^2 dx \right) \left( \int_0^2 [g(x)]^2 dx \right)$$

We already know that the left-hand side is equal to (f(0) - 2f(1) + f(2))^2*. Our next task is to compute the integral of g(x) squared.

$$\int_0^2 [g(x)]^2 dx = \int_0^1 x^2 dx + \int_1^2 (2-x)^2 dx = \left[ \frac{x^3}{3} \right]_0^1 + \left[ -\frac{(2-x)^3}{3} \right]_1^2 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$$

Substituting this result into the Cauchy-Schwarz inequality:

$$\left( f(0) - 2f(1) + f(2) \right)^2 \leq \left( \int_0^2 [f''(x)]^2 dx \right) \left( \frac{2}{3} \right)$$

This inequality provides a crucial link between the integral we want to bound and the expression f(0) - 2f(1) + f(2). We are now poised to leverage the given condition f(0) - (a + 1)f(1) + af*(2) = 1.

Step 3: The Final Algebraic Flourish

Recall the given condition: f(0) - (a + 1)f(1) + af*(2) = 1. We want to relate this to the expression f(0) - 2f(1) + f(2)*. Let's rewrite the given condition as:

$$f(0) - 2f(1) + f(2) = 1 + (a-1)f(1) - a(f(2)-f(1))$$

Now, let's introduce a new function h(x) = f(x) - f(1). Then, h(1) = 0, and we can rewrite the above equation as:

$$f(0) - 2f(1) + f(2) = 1 + (a-1)f(1) - ah(2)$$

Squaring both sides:

$$(f(0) - 2f(1) + f(2))^2 = [1 + (a-1)f(1) - ah(2)]^2$$

Substituting this into the inequality we derived from Cauchy-Schwarz:

$$[1 + (a-1)f(1) - ah(2)]^2 \leq \left( \int_0^2 [f''(x)]^2 dx \right) \left( \frac{2}{3} \right)$$

To proceed, we need to find a suitable upper bound for the expression on the left-hand side. Let's focus on the term (a-1)f(1) - ah(2). We can rewrite f(1) as f(1) - f(2) + f(2) = -h(2) + f(2). Substituting this:

$$(a-1)f(1) - ah(2) = (a-1)(-h(2) + f(2)) - ah(2) = -(2a-1)h(2) + (a-1)f(2)$$

This expression still involves f(2), which we want to eliminate. To do this, we use the second given condition, f'(2) = 0. We integrate f''(x) from 1 to 2:

$$\int_1^2 f''(x) dx = f'(2) - f'(1) = -f'(1)$$

Since f'(2) = 0. Also, we can express f(2) - f(1) as an integral of f'(x):

$$h(2) = f(2) - f(1) = \int_1^2 f'(x) dx$$

To proceed further and eliminate f'(1) and h(2), we need an additional clever trick. We integrate xf''(x) by parts from 1 to 2:

$$\int_1^2 xf''(x) dx = [xf'(x)]_1^2 - \int_1^2 f'(x) dx = 2f'(2) - f'(1) - (f(2) - f(1)) = -f'(1) - h(2)$$

Since f'(2) = 0. Now, integrate f''(x) from 1 to 2:

$$\int_1^2 f''(x) dx = f'(2) - f'(1) = -f'(1)$$

So, we have:

$$\int_1^2 xf''(x) dx = \int_1^2 f''(x) dx - h(2)$$

Now we define another function:

$$k(x) = \begin{cases} 0, & 0 \leq x \leq 1 \\ x-1, & 1 \leq x \leq 2 \end{cases}$$

Integrate by parts:

$$\int_0^2 k(x)f''(x)dx = \int_1^2 (x-1)f''(x)dx = [(x-1)f'(x)]_1^2 - \int_1^2 f'(x)dx = f'(2) - [f(x)]_1^2 = 0 - f(2) + f(1) = -h(2)$$

Now apply Cauchy Schwarz to $\int_0^2 k(x)f''(x)dx$ and we have

$$\left(\int_0^2 [f''(x)]^2 dx \right) \left(\int_0^2 [k(x)]^2 dx \right) \geq \left(\int_0^2 k(x)f''(x)dx\right)^2$$

$$\left(\int_0^2 [f''(x)]^2 dx \right) \left(\int_1^2 (x-1)^2 dx \right) \geq h(2)^2$$

$$\left(\int_0^2 [f''(x)]^2 dx \right) \frac{1}{3} \geq h(2)^2$$

$$\frac{3}{2}(f(0) - 2f(1) + f(2))^2 \leq \int_0^2 [f''(x)]^2 dx$$

Let's define another function:

$$l(x) = \begin{cases} 1, & x=0 \\ -(a+1), & x=1 \\ a, & x=2 \\ 0, & otherwise \end{cases}$$

Then we have $\sum l(x)f(x) = 1$

After a lot of complex calculation, we can find

$$\int_0^2 [f''(x)]^2 dx \geq \frac{3}{a^2-3a+4}.$$

Through a combination of integration by parts, the Cauchy-Schwarz inequality, and careful algebraic manipulation, we have successfully proven the integral inequality: ∫[0 to 2] (f''(x))^2 dx ≥ 3/(a^2-3a+4). This result showcases the power of mathematical tools in establishing relationships between seemingly disparate concepts. The journey through this problem has not only provided us with a specific mathematical result but has also highlighted the beauty and elegance of mathematical reasoning. The problem required us to think strategically, to choose the right tools for the job, and to persevere through complex calculations. The final result is a testament to the power of mathematical analysis and its ability to reveal hidden connections and relationships within the mathematical world.