If I have a sum utility function like this $U(y) = \sum_{j=1}^J u(y_j)$ where $u$ is convex.
Is there then a way to find walrasian demand for such a function without using calculus or do you need to use lagrange to solve for j,...,J?
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Given that the utility maximization problem is: \begin{eqnarray*} \max_{\{y_j \geq 0 : 1\leq j\leq J\}} && \sum_{j=1}^{J} u(y_j) \\ \text{s.t.} && \sum_{j=1}^{J} p_jy_j \leq m\end{eqnarray*} where $p_j > 0$ for every $j$, and $m > 0$.
If $u$ is convex and strictly increasing, then in optimum consumer will always spend all his money on the cheapest good. That is, buying $\frac{m}{p_1}$ units of commodity 1 when $p_1 \leq \min_j p_j$ will be utility maximizing for the consumer.
Here is the proof: First of all note that in optimum consumer will never spend less than his income because his utility is increasing. So, we are left with bundles on which he spends all his income. Now any consumption bundle that costs exactly equal to his income can be written as the convex combination of the bundles of the type in which the consumer spends all his money on a single good.
For example, in a two commodity case: Any $(y_1, y_2)$ on the budget line can be written as the convex combination of $\left(\frac{m}{p_1}, 0\right)$ and $\left(0,\frac{m}{p_2}\right)$.
Since, $u$ is convex, and the sum of convex functions is convex, so we also know that the utility $U$ is convex. By convexity of $U$, utility at the convex combination of the bundles of the type in which the consumer spends all his money on a single good is less than or equal to the convex combination of the utilities at those bundles, and that is in turn less than or equal to the maximum of the utilities at those bundles. This shows that the consumer can never lose by changing his choice to the bundle where he buys the cheapest bundle.
For example: In the two commodity case, we can write any bundle any bundle $(y_1, y_2)$ on the budget line as $\lambda\left(\frac{m}{p_1}, 0\right)+(1-\lambda)\left(0,\frac{m}{p_2}\right)$ where $0 \leq \lambda \leq 1$. Therefore, by convexity of $U$,
\begin{eqnarray*} U(y_1, y_2) = U\left(\lambda\left(\frac{m}{p_1}, 0\right)+(1-\lambda)\left(0,\frac{m}{p_2}\right)\right) & \leq & \lambda U\left(\frac{m}{p_1}, 0\right)+(1-\lambda)U\left(0,\frac{m}{p_2}\right) \\ & \leq & \max\left( U\left(\frac{m}{p_1}, 0\right),U\left(0,\frac{m}{p_2}\right)\right)\end{eqnarray*} This implies that the utility is maximum at either $\left(\frac{m}{p_1}, 0\right)$ or $\left(0,\frac{m}{p_2}\right)$. Given that $U(y_1, y_2) = u(y_1)+u(y_2)$, it follows that it is maximum when one consumes the cheaper one.
