We are given that $f: \mathbb{R} \to \mathbb{R}$ satisfies the functional equation

$f(x+1) = f(x) \quad \text{for all} \quad x \in \mathbb{R},$

and that $f$ is integrable over the interval $[0, 1]$. Define the function $F(x)$ as the integral of $f$ from 0 to $x$, i.e.,

$F(x) := \int_0^x f(t) \, dt.$

We are asked to prove that

$F(x + 1) = F(x) \quad \text{for all} \quad x \in \mathbb{R} \quad \Longleftrightarrow \quad \int_0^1 f(t) \, dt = 0.$

Step 1: Prove the forward direction: $F(x + 1) = F(x) \implies \int_0^1 f(t) \, dt = 0$

1.1 First, let's consider the condition $F(x + 1) = F(x)$ for all $x \in \mathbb{R}$. In particular, this must hold for integer values of $x$.

1.2 For $x = 0$, we have:

$F(1) = F(0).$

By the definition of $F(x)$, we have:

$F(1) = \int_0^1 f(t) \, dt \quad \text{and} \quad F(0) = \int_0^0 f(t) \, dt = 0.$

Thus, we have:

$\int_0^1 f(t) \, dt = 0.$

This completes the forward direction.

Step 2: Prove the reverse direction: $\int_0^1 f(t) \, dt = 0 \implies F(x + 1) = F(x)$ for all $x \in \mathbb{R}$

2.1 Now assume that $\int_0^1 f(t) \, dt = 0$. We want to show that $F(x + 1) = F(x)$ for all $x \in \mathbb{R}$.

2.2 Let $x = n + \alpha$, where $n \in \mathbb{Z}$ (an integer) and $\alpha \in [0, 1)$. Then, using the definition of $F(x)$, we have:

$F(x + 1) = F(n + \alpha + 1) = \int_0^{n + \alpha + 1} f(t) \, dt.$

We can split the integral as follows:

$\int_0^{n + \alpha + 1} f(t) \, dt = \int_0^n f(t) \, dt + \int_n^{n + \alpha} f(t) \, dt + \int_{n + \alpha}^{n + \alpha + 1} f(t) \, dt.$

Since $f(x + 1) = f(x)$ for all $x$, we know that $f(t)$ is periodic with period 1. Therefore, we can substitute $f(t) = f(t - 1)$ in the last integral:

$\int_{n + \alpha}^{n + \alpha + 1} f(t) \, dt = \int_0^1 f(t) \, dt = \int_0^1 f(t) \, dt.$

Thus, we have:

$F(x + 1) = \int_0^n f(t) \, dt + \int_n^{n + \alpha} f(t) \, dt + \int_0^1 f(t) \, dt.$

Now, since $\int_0^n f(t) \, dt$ is just the sum of the integrals over the intervals $[0, 1], [1, 2], \dots, [n-1, n]$, it follows that:

$\int_0^n f(t) \, dt = n \cdot \int_0^1 f(t) \, dt.$

So,

$F(x + 1) = n \cdot \int_0^1 f(t) \, dt + \int_n^{n + \alpha} f(t) \, dt + \int_0^1 f(t) \, dt.$

Now, using the same substitution $\int_0^1 f(t) \, dt = 0$, we conclude:

$F(x + 1) = F(x).$

Thus, $F(x + 1) = F(x)$ for all $x \in \mathbb{R}$, as required.

Conclusion

We have shown both directions:

$F(x + 1) = F(x) \quad \Longleftrightarrow \quad \int_0^1 f(t) \, dt = 0.$

Let me know if you'd like further clarification or have any questions on the steps!

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Related Questions

1. How does the periodicity of $f(x)$ affect the behavior of $F(x)$?
2. What happens if we change the interval of integration, say from $[a, b]$ to $[a+1, b+1]$?
3. How can we generalize the result for functions that are periodic with period $p$ (i.e., $f(x + p) = f(x)$)?
4. What would happen if $f(x)$ were not integrable over $[0, 1]$?
5. What is the significance of the condition $f(x + 1) = f(x)$ in applications like signal processing or Fourier analysis?

Tip: The periodicity condition $f(x + 1) = f(x)$ often leads to useful simplifications in integrals, especially when dealing with periodic functions and their integrals over specific intervals.