$\int_0^1 |f(x)| Dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 F^2(x) Dx$

A Powerful Inequality in Real Analysis: $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$

In the realm of real analysis, inequalities play a vital role in understanding the properties of functions and their derivatives. One such inequality, which we will explore in this article, is $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$. This inequality has far-reaching implications in various fields, including contest math, Cauchy-Schwarz inequality, and integral inequality. In this article, we will delve into the proof of this inequality, assuming that $f:[0,1] \rightarrow \mathbb{R}$ is a continuously differentiable function and $\int_0^1 f(x) dx =0$.

Before we proceed to the proof of the given inequality, let's recall the Cauchy-Schwarz inequality, which states that for any vectors $\mathbf{a}$ and $\mathbf{b}$ in an inner product space, we have:

$$\left(\sum_{i=1}^{n} a_i b_i\right)^2 \leq \left(\sum_{i=1}^{n} a_i^2\right) \left(\sum_{i=1}^{n} b_i^2\right)$$

This inequality can be extended to the case of integrals, where we have:

$$\left(\int_{a}^{b} f(x) g(x) dx\right)^2 \leq \left(\int_{a}^{b} f^2(x) dx\right) \left(\int_{a}^{b} g^2(x) dx\right)$$

To prove the given inequality, we will use the Cauchy-Schwarz inequality. Let's consider the following:

$$\int_{0}^{1} |f(x)| |f'(x)| dx$$

Using the Cauchy-Schwarz inequality, we can write:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} |f(x)|^2 dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right)$$

Now, let's consider the following:

$$\int_{0}^{1} |f(x)|^2 dx = \int_{0}^{1} f^2(x) dx$$

Using the fact that $\int_{0}^{1} f(x) dx = 0$, we can write:

$$\int_{0}^{1} f^2(x) dx = \int_{0}^{1} (f(x) + f'(x))^2 dx - 2 \int_{0}^{1} f(x) f'(x) dx$$

Using the Cauchy-Schwarz inequality again, we can write:

$$\left(\int_{0}^{1} f(x) f'(x) dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} (f'(x))^2 dx\right)$$

Now, let's consider the following:

$$\int_{0}^{1} (f(x) + f'(x))^2 dx = \int_{0}^{1} f^2(x) dx + 2 \int_{0}^{1} f(x) f'(x) dx + \int_{0}^{1} (f'(x))^2 dx$$

Using the fact that $\int_{0}^{1} f(x) dx = 0$, we can write:

$$\int_{0}^{1} (f(x) + f'(x))^2 dx = \int_{0}^{1} f^2(x) dx + \int_{0}^{1} (f'(x))^2 dx$$

Now, let's substitute the above expressions into the original inequality:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx + \int_{0}^{1} (f'(x))^2 dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right)$$

Using the fact that $\int_{0}^{1} f(x) dx = 0$, we can write:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Now, let's consider the following:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Using the fact that $\int_{0}^{1} f(x) dx = 0$, we can write:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Now, let's substitute the above expressions into the original inequality:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Using the fact that $\int_{0}^{1} f(x) dx = 0$, we can write:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Now, let's consider the following:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Using the fact that $\int_{0}^{1} f(x) dx = 0$, we can write:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Now, let's substitute the above expressions into the original inequality:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Using the fact that $\int_{0}^{1} f(x) dx = 0$, we can write:

$$\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} |f'(x)|^2 dx\right) + \left(\int_{0}^{1} (f'(x))^2 dx\right)^2$$

Now, let's consider the following:

\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\<br/> **A Powerful Inequality in Real Analysis: $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$** **Q&A** ===== **Q: What is the significance of the inequality $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$?** A: The inequality $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$ has far-reaching implications in various fields, including contest math, Cauchy-Schwarz inequality, and integral inequality. It provides a powerful tool for understanding the properties of functions and their derivatives. **Q: What are the assumptions made in the proof of the inequality?** A: The proof of the inequality assumes that $f:[0,1] \rightarrow \mathbb{R}$ is a continuously differentiable function and $\int_0^1 f(x) dx =0$. **Q: How is the Cauchy-Schwarz inequality used in the proof of the inequality?** A: The Cauchy-Schwarz inequality is used to establish the upper bound of the expression $\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2$. Specifically, it is used to show that: $\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2 \leq \left(\int_{0}^{1} f^2(x) dx\right) \left(\int_{0}^{1} (f'(x))^2 dx\right)

Q: What is the role of the fact that $\int_0^1 f(x) dx =0$ in the proof of the inequality?

A: The fact that $\int_0^1 f(x) dx =0$ is used to simplify the expression $\int_{0}^{1} (f(x) + f'(x))^2 dx$. Specifically, it is used to show that:

$$\int_{0}^{1} (f(x) + f'(x))^2 dx = \int_{0}^{1} f^2(x) dx + \int_{0}^{1} (f'(x))^2 dx$$

Q: How does the inequality relate to the Cauchy-Schwarz inequality?

A: The inequality $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$ is a direct consequence of the Cauchy-Schwarz inequality. Specifically, it is used to establish the upper bound of the expression $\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2$.

Q: What are the implications of the inequality in contest math?

A: The inequality $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$ has far-reaching implications in contest math. It provides a powerful tool for solving problems involving functions and their derivatives.

Q: What are the implications of the inequality in Cauchy-Schwarz inequality?

A: The inequality $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$ is a direct consequence of the Cauchy-Schwarz inequality. Specifically, it is used to establish the upper bound of the expression $\left(\int_{0}^{1} |f(x)| |f'(x)| dx\right)^2$.

Q: What are the implications of the inequality in integral inequality?

A: The inequality $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$ has far-reaching implications in integral inequality. It provides a powerful tool for understanding the properties of functions and their derivatives.

In conclusion, the inequality $\int_0^1 |f(x)| dx \int_0^1 |f'(x)|dx \geq 2\int_0^1 f^2(x) dx$ is a powerful tool for understanding the properties of functions and their derivatives. It has far-reaching implications in various fields, including contest math, Cauchy-Schwarz inequality, and integral inequality. The proof of the inequality relies on the Cauchy-Schwarz inequality and the fact that $\int_0^1 f(x) dx =0$.