Find a function `g:RrarrR`, continuous in `[0,oo)` and positive in `(0,oo)` satisfying `g(1)=1` and `1/2int_0^xg^2(t)dt=1/x(int_0^xg(t)dt)^2`

Answer

To solve the problem, we need to find a function \( g: \mathbb{R} \to \mathbb{R} \) that is continuous on \([0, \infty)\), positive on \((0, \infty)\), satisfies \( g(1) = 1 \), and fulfills the equation:

\[ \frac{1}{2} \int_0^x g^2(t) \, dt = \frac{1}{x} \left( \int_0^x g(t) \, dt \right)^2 \]

### Step 1: Define the Function
Let \( F(x) = \int_0^x g(t) \, dt \). Then, by the Fundamental Theorem of Calculus, we have:

\[ F'(x) = g(x) \]

### Step 2: Substitute \( F(x) \) into the Given Equation
Substituting \( F(x) \) into the equation, we rewrite it as:

\[ \frac{1}{2} \int_0^x (F'(t))^2 \, dt = \frac{1}{x} (F(x))^2 \]

### Step 3: Differentiate Both Sides
Differentiating both sides with respect to \( x \):

Using the Leibniz rule on the left side:

\[ \frac{1}{2} (F'(x))^2 = \frac{d}{dx} \left( \frac{1}{x} (F(x))^2 \right) \]

Using the product rule on the right side:

\[ \frac{d}{dx} \left( \frac{1}{x} (F(x))^2 \right) = -\frac{1}{x^2} (F(x))^2 + \frac{2 F(x) F'(x)}{x} \]

### Step 4: Set the Derivatives Equal
Setting the derivatives equal gives us:

\[ \frac{1}{2} (F'(x))^2 = -\frac{1}{x^2} (F(x))^2 + \frac{2 F(x) F'(x)}{x} \]

### Step 5: Rearranging the Equation
Rearranging the equation, we have:

\[ \frac{1}{2} (F'(x))^2 - \frac{2 F(x) F'(x)}{x} + \frac{1}{x^2} (F(x))^2 = 0 \]

### Step 6: Recognizing the Quadratic Form
This is a quadratic equation in \( F'(x) \):

\[ \frac{1}{2} (F'(x))^2 - \frac{2 F(x)}{x} F'(x) + \frac{1}{x^2} (F(x))^2 = 0 \]

### Step 7: Solving the Quadratic Equation
Using the quadratic formula \( F'(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = \frac{1}{2}, b = -\frac{2 F(x)}{x}, c = \frac{1}{x^2} (F(x))^2 \).

Calculating the discriminant:

\[ b^2 - 4ac = \left(-\frac{2 F(x)}{x}\right)^2 - 4 \cdot \frac{1}{2} \cdot \frac{1}{x^2} (F(x))^2 = \frac{4 F(x)^2}{x^2} - \frac{2 F(x)^2}{x^2} = \frac{2 F(x)^2}{x^2} \]

Thus, we have:

\[ F'(x) = \frac{2 F(x)}{x} \pm \sqrt{\frac{2 F(x)^2}{x^2}} = \frac{2 F(x)}{x} \pm \frac{\sqrt{2} F(x)}{x} \]

### Step 8: Simplifying the Expression
This gives us two potential solutions:

1. \( F'(x) = \frac{(2 + \sqrt{2}) F(x)}{x} \)
2. \( F'(x) = \frac{(2 - \sqrt{2}) F(x)}{x} \)

### Step 9: Solving the Differential Equation
Solving the first differential equation:

\[ \frac{F'(x)}{F(x)} = \frac{(2 + \sqrt{2})}{x} \]

Integrating both sides:

\[ \ln |F(x)| = (2 + \sqrt{2}) \ln |x| + C \]

Thus,

\[ F(x) = C x^{2 + \sqrt{2}} \]

### Step 10: Finding \( g(x) \)
Since \( g(x) = F'(x) \):

\[ g(x) = C (2 + \sqrt{2}) x^{1 + \sqrt{2}} \]

### Step 11: Applying the Condition \( g(1) = 1 \)
To satisfy \( g(1) = 1 \):

\[ C (2 + \sqrt{2}) = 1 \implies C = \frac{1}{2 + \sqrt{2}} \]

### Final Function
Thus, the function \( g(x) \) is:

\[ g(x) = \frac{1}{2 + \sqrt{2}} (2 + \sqrt{2}) x^{1 + \sqrt{2}} = x^{1 + \sqrt{2}} \]

### Conclusion
The function \( g(x) = x^{1 + \sqrt{2}} \) is continuous on \([0, \infty)\) and positive on \((0, \infty)\), satisfying \( g(1) = 1 \).