I came across the following problem:
Find the smallest value of $x_1^2 + x_2^2 + x_3^2 + x_4^2$ where $x_1 - x_2 + 2x_3 - x_4 = 1$.
Setting $g(x) = x_1 - x_2 + 2x_3 - x_4 - 1$ and using Lagrange multipliers $(\nabla f) = \lambda(\nabla g)$ you can easily find $f(1/7, -1/7, 2/7, -1/7) = 1/7$ to be a local extrema.
For the final step which is proving that this point is in fact a global minimum, the solution provided by my professor states that:
Since $f$ can assume arbitrarily large values, for example along the line $x = (1 + t, t, 0, 0)$ as $t \rightarrow \infty$ and is non-negative, our candidate is a minimum value.
I understand this proves that the point cannot be a global maxima. However I must humbly ask: Is this sufficient proof that this point is a global minima?
(i) How can we know that it is not a local maxima?
(ii) How can we know that the function can't assume arbitrarily small values elsewhere? $f$ is non-negative and has no other extrema, however can I be sure there is no limit where $f$ tends to a smaller value than $1/7$, without further investigation?
This particular case can be solved by direct substitution so defining
$$ f(x_1,x_2,x_3) = x_1^2+x_2^2+x_3^2+(x_1-x_2+2x_3-1)^2 $$
and calculating the Hessian at the stationary point $X^* = (\frac 17,-\frac 17, \frac 27)$ we have
$$ H(X^*) = \left( \begin{array}{ccc} 4 & -2 & 4 \\ -2 & 4 & -4 \\ 4 & -4 & 10 \\ \end{array} \right) $$
which is positive definite so the found stationary point is a local minimum.