Newton's method - is ascent guaranteed if the function is concave?

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Suppose we have a concave multivariate function $L(\theta):\mathbb{R}^n \to \mathbb{R}$.

We want to use Newton's method to maximise $L(\theta)$ by using the iteration:

$$\theta_{n+1} = \theta_n - [d^2 L(\theta_n)]^{-1}\nabla L(\theta_n)$$

One thing I am struggling to understand is how the concavity of $L(\theta)$ affects the problems associated with Newton's method.

For example, two problems pointed out in my textbook are:

  1. Inversion of $d^2 L(\theta_n)$ at each iteration may be computationally expensive

  2. Newton's method is not guaranteed to ascent $L(\theta)$ at any given iteration.

Now, the concavity does not affect point $(1)$, however I am wondering if concavity actually does affect point $(2)$ by guaranteeing that each iteration does ascend $L(\theta)$?

I am not sure how to justify if it does or does not, and would appreciate a bit of insight as to why it does or does not.

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Let $f:\mathbb R\to\mathbb R$ be defined piecewise: If $x\leq-1$ then $f(x)=-x^2$, and if $x\geq-1$ then $f(x)=-1-9(x+1)$. In other words, on the left side, $f$ locally looks nice and predictable; but $f$ drops off a cliff without warning.

Assume that I did the algebra right, and $f$ is a concave function...

If our starting point is $x_0=-2$ then the next iterate from Newton's method is $x_1=0$, overshooting the argmax. Worse, our objective $f(x_0)=-4$ has decreased to $f(x_1)=-10$.