Proof $ f(x,y) = |x| + |y| $ has local minimum at $(0,0)$

Question

I'm Trying in various ways but failing , As :

$ \nabla f(x,y) = 0 $ gives me:

$f_x = \frac{x}{|x|} = 0;\quad f_y = \frac{y}{|y|} = 0,$ and both cannot be solved as assuming:

$x=0 \rightarrow \frac{0}{|0|} = \mathrm{ undetermined }$ and the same goes for $y=0$.

I'd appreciate any kind of help.

Answer 1

First, recall that $|x| \ge 0$ for all $x$. I suspect that you are allowed to assume this, without having to prove it. Then, $$f(x, y) = |x| + |y| \ge |x| + 0 \ge 0 + 0 = f(0, 0)$$ for all $x, y$. This proves that $f$ is a global minimum (not just a local minimum). That's all the proof that's necessary.

As others have pointed out in the comments, taking derivatives is a dead end. Derivatives are an extremely useful tool in finding maxima and minima, but the derivative must exist in order for us to use it. It is perfectly possible to have a function (like this one) that achieves a minimum at a sharp point (or sharp edge), in which case the derivative will not exist. In these cases, other methods, including ad-hoc proofs like the one above, are necessary to show a function has a minimum/maximum.

Answer 2

The absolute value is always non-negative

$$|x| = 0 \quad\Longleftrightarrow \quad x=0$$ Thus $$f(x,y) = 0 \quad\Longleftrightarrow \quad x=y=0$$ $$f(x,y) > 0 \quad\Longleftrightarrow \quad x\neq0 \lor y\neq0$$ This means $f$ has s global minimum at $(0,0)$, thus there is also a local minimum.

Answer 3

$|x|+|y|\geq 0$ for all $(x,y)$. Then $f(x,y)\geq0 =f(0,0)$ for all $(x,y)$ and since $f(x,y)$ is clearly convex, local minimum is global minimum! We are done!!! If you are familiar with subdifferentials of convex analysis you can use them, but you cannot use differentials!