Let we are given with a function of infinite variables $$f(x_1, x_2, \ldots) $$ which is differentiable. Now if we know that minimum of the above function exist, then can we apply the same condition i.e. $$\frac{\partial f}{\partial x_i} = 0 \qquad \text{for each} \ i$$ to find out the critical points even in the case of infinite variables?
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Yes, because the function may be viewed as a function of just the variable $x_i$ with all the remaining arguments fixed.
Define the "fixed" function $f_{x_i}$ at a point $z = (z_1, z_2, \dots)$, to be $f_{x_i} : \mathbb{R} \to \mathbb{R}$ which has $f_{x_i}(x) = f(z_1, z_2, \dots, z_{i-1}, x, z_{i+1}, \dots)$.
Suppose $z$ is a stationary point of $f$. Then $z_i$ is a stationary point of $f_{x_i}$.
So if we observe that each $f_{x_i}'(z_i) = 0$, then $(z_1, z_2, \dots)$ is a candidate critical point of $f$. I think the converse is true (that is, if the partial derivatives are all 0 then the derivative is 0), but I've not got much time to remember the proof right now.
Patrick Stevens
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Can you please check this? https://math.stackexchange.com/questions/2504165/to-find-out-the-minimum-of-following-convex-functional. Thanks – PAMG Nov 05 '17 at 09:32