Gradient of function $\nabla g(x)$ when $g = f^2$

Question

I can't solve this question. Someone could explain for me please?

Consider $f: R^n \rightarrow R$ differentiable function. Define for all $x \in R^n$, the functions: $$g(x) = \max \{ 0, f(x) \}\;\; \mbox{and}\;\; h(x) = g(x)^2.$$ So, $\nabla h(x) = 2*g(x)*\nabla f(x).$

I can't solve this.

I am thinking this problem has a direct solution, but I can't solve it. I was thinking this way:

How $f$ is differentiable, then $g$ and $h$ are differentiable too. And for $x \in R^n$ when $f(x) \leq 0$, we have $g(x) = 0$ and the result is true because $\nabla h(x) = 0$.

The problem for me is when $f(x) > 0$. I know that for $x$ such that $f(x) > 0$, we have $g(x) = f(x)$ and $h(x) = f(x)^2$.

But I couldn't to prove the result. The result remember me the chain rule differentiable, but I didn't find a chain rule differentiable for gradients vectors.

Answer 1

You don't need an explicit "chain rule" for two reasons. First, "chain rule"s apply to composition of functions; you want a product rule. Second, the derivation is actually not that hard. Remember the gradient of $h(x)^2$ is gotten by taking partial derivatives: $\frac{\partial}{\partial x_i} h(x)^2$ and you can do this by hand.

Actually if $f$ is differentiable, then $g$ might not be differentiable, because it might have a "kink" if $f$ has both positive and negative values. You would further have to explain how $h$ is differentiable.

Note: The calculations should be fine where $f(x) < 0$ or $f(x) > 0$, but the problems with differentiability happen when $f(x) = 0$.

Example: $f(x) = x$ from $\mathbb{R} \to \mathbb{R}$. Then $g(x) = 0$ when $x < 0$ and $g(x) = x$ when $x \geq 0$, which has a kink at $0$. However, when we define $h(x) = g(x)^2$, we see that $h(x) = 0$ when $x < 0$, but $h(x) = x^2$ when $x \geq 0$ and so there is no kink in $h$ and the derivative of $h$ is in fact $2g(x)f'(x)$.

Answer 2

With your tips, I thinking this problem in this way.

I can write $g(x) = \dfrac{f(x) + |f(x)|}{2}$ and then $h(x) = \dfrac{f(x)^2 + f(x) | f(x) |}{4}$.

So, now I will calculate by hand $\dfrac{\partial}{\partial x_i} h(x)$. For to simply my idea, first all, I will calculate this derivative (thinking in $\mathbb{R}$): $$ \displaystyle\lim_{x \rightarrow x_0} \dfrac{h(x) - h(x_0)}{x-x_0} = \displaystyle\lim_{x \rightarrow x_0} \dfrac{1}{2} \left( \dfrac{(f(x) + f(x_0))(f(x) - f(x_0))}{x-x_0} + \dfrac{f(x) | f(x)| - f(x_0)|f(x_0)|}{x-x_0} \right) . $$

If I pass the limit, we have in the first parcel $\dfrac{1}{2} (2f(x_0)f'(x_0) + A)$, where $A = \dfrac{f(x) | f(x)| - f(x_0)|f(x_0)|}{x-x_0}$.

Now, I need to show that $A = 2 | f(x_0)| f'(x_0)|$. But I can't do it.