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I want to prove the convergence of the newton method for a convex function.

Let $f: [a,b] \to \mathbb R$ be a differentiable, convex function with $f(a) \gt 0$ and $f(b) \lt 0$. Let $(x_n)_n$ be the sequence

$$x_0 = a, \quad x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

Show that $(x_n)_n$ converges to the unique zero of the function $f$.

I understand the geometric intuition with $(x_n)_n$ being the sequence of zeros of the tangent lines of the function. But I struggle proving it mathematically.

I already know, that $f$ has a unique zero $c \in (a,b)$. Proceeding from there I want to show that if $(x_n)_n$ converges, then $\lim_{n \to \infty} (x_n)_n = c$ and also that $(x_n)_n$ is increasing and bounded. Then it follows that $\lim_{n \to \infty} (x_n)_n = c$.

I found this question but the proof restricts itself to different cases and uses just geometric arguments for the increasingness and boundness of $(x_n)_n$. I'm worried that this won't be rigorously enough.

How can I prove the above theorem? I tried proving the first statement below.

What I have so far:

Lemma 1: If $(x_n)_n$ converges then it converges to the zero of $f$.

Assume $(x_n)_n$ converges with $\lim_{n \to \infty} x_n = c$. Then it follows that $\lim_{n \to \infty} (x_n - x_{n+1}) = 0$. Also $f'$ is bounded since it's monotonously increasing on a compact interval. Therefore

$$0 = \lim_{n \to \infty} f'(x_n) (x_n - x_{n+1}) = \lim_{n \to \infty} f(x_n) = f(c)$$

where the last equality follows from the continuity of $f$.

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mdcq
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2 Answers2

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In this situation, f decreases through its zero. Assuming convexity, you can show geometrically that Newton will undershoot if you start left of the zero and overshoot if you start right of the zero (and left of the minimum, which is not guaranteed by your assumptions). This is because f lies above its tangent lines. As a result you have at most one step to get to the left of the zero, after which you have a bounded increasing sequence.

Ian
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  • Thanks! Could you elaborate how to then show that the sequence is increasing and bounded? – mdcq Jan 22 '18 at 19:04
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    @philmcole It's just induction: you're on the left, you undershoot, and then you're on the left again, so you undershoot again, etc. ad infinitum. – Ian Jan 22 '18 at 22:14
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f is differentiable and convex , $f(c)=0$ .If $A(x_0,f(x_0))$ is a point of $C_f$ with $x_0 <c$ then the tangent at $x=x_0$ is $(e):y-f(x_0)=f'(x_0)(x-x_0)$ for $y=0:$ $$x-x_0= -\frac{f(x_0)}{f'(x_0)}$$ and because $f$ is convex then $(e)$ is bellow $C_f$

So $x_0<x<c$

For the sequence we have ${x_0}=a<c$ as a result of applying the above $x_0<x_1<c$ with induction we get $x_0<x_1<x_2<...<x_n$ So the sequence converges. So $lim{n\to +\infty}({x_{n+1}-x_n})=0\Rightarrow lim{n\to +\infty}(-\frac{f(x_n)}{f'(x_n)})=0 $

We know that $f'$ is continious because it is increasing and not zero $\forall x\leq c$

So $ lim{n\to +\infty} (f(x_n))=0$ so $x_n\rightarrow c$

Lerigorilla
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