For a real-valued function $f(x,y)$ of two positive real variables $x$ and $y$, define $f$ to be [i]linearly bounded[/i] if and only if there exists a positive number $K$ such that $|f(x,y)| < K(x+y)$ for all positive $x$ and $y.$ Find necessary and sufficient conditions on the real numbers $\alpha$ and $\beta$ such that $x^{\alpha}y^{\beta}$ is linearly bounded.
To determine the necessary and sufficient conditions on the real numbers $\alpha$ and $\beta$ such that $x^{\alpha}y^{\beta}$ is linearly bounded, we need to analyze the behavior of the function $f(x, y) = x^{\alpha}y^{\beta}$ under the given condition $|f(x, y)| < K(x + y)$ for some positive constant $K$ and for all positive $x$ and $y$.

1. **Suppose $x^{\alpha}y^{\beta}$ is linearly bounded.**
   - This means there exists a constant $K > 0$ such that $|x^{\alpha}y^{\beta}| < K(x + y)$ for all $x, y > 0$.

2. **Consider the case $\alpha + \beta > 1$.**
   - Let $x = y \to \infty$. Then $x^{\alpha}y^{\beta} = x^{\alpha + \beta}$.
   - If $\alpha + \beta > 1$, then $x^{\alpha + \beta} \to \infty$ as $x \to \infty$.
   - However, $K(x + y) = 2Kx \to \infty$ as $x \to \infty$.
   - This does not lead to a contradiction, so we need to refine our argument.

3. **Consider the case $\alpha + \beta < 1$.**
   - Let $x = y \to 0$. Then $x^{\alpha}y^{\beta} = x^{\alpha + \beta}$.
   - If $\alpha + \beta < 1$, then $x^{\alpha + \beta} \to 0$ as $x \to 0$.
   - However, $K(x + y) = 2Kx \to 0$ as $x \to 0$.
   - This does not lead to a contradiction, so we need to refine our argument.

4. **Consider the case $\alpha < 0$.**
   - Let $x \to 0$ and $y = 1$. Then $x^{\alpha}y^{\beta} = x^{\alpha}$.
   - If $\alpha < 0$, then $x^{\alpha} \to \infty$ as $x \to 0$.
   - However, $K(x + y) = K(x + 1) \to K$ as $x \to 0$.
   - This leads to a contradiction because $x^{\alpha}$ grows unbounded while $K(x + 1)$ remains bounded.

5. **Consider the case $\beta < 0$.**
   - Let $y \to 0$ and $x = 1$. Then $x^{\alpha}y^{\beta} = y^{\beta}$.
   - If $\beta < 0$, then $y^{\beta} \to \infty$ as $y \to 0$.
   - However, $K(x + y) = K(1 + y) \to K$ as $y \to 0$.
   - This leads to a contradiction because $y^{\beta}$ grows unbounded while $K(1 + y)$ remains bounded.

6. **Consider the case $\alpha + \beta = 1$ and $\alpha, \beta \geq 0$.**
   - By Young's inequality, for any $\alpha, \beta \geq 0$ such that $\alpha + \beta = 1$, we have:
     \[
     x^{\alpha}y^{\beta} \leq \alpha x + \beta y.
     \]
   - Therefore, $x^{\alpha}y^{\beta}$ is linearly bounded by $K(x + y)$ with $K = \max(\alpha, \beta)$.

Thus, the necessary and sufficient conditions for $x^{\alpha}y^{\beta}$ to be linearly bounded are $\alpha + \beta = 1$ and $\alpha, \beta \geq 0$.

The final answer is $\boxed{\alpha + \beta = 1, \alpha, \beta \geq 0}$.