$f(x+\lambda p) \le f(x) + c\lambda p^t\nabla f(x)$ then there's always $0<\lambda_1<\lambda$ that also satisfies this?

Question

Let $f:\mathbb{R}^n\to\mathbb{R}, x, d\in\mathbb{R}^n$ and $\lambda>0$ such that $x+\lambda d$ satisfies Armijo's condition. Let $0<\lambda_1<\lambda$. Does $\lambda_1$ satisfies Armijo too? Prove or give a counter example.

The armijo condition can be written as this:

$$f(x+\lambda p) \le f(x) + c\lambda p^t\nabla f(x)$$

for some $c\in(0,1)$. I guess it is asking that: if the condition is met for some $c$ and $\lambda$, then it is met by another $c_1$ and $\lambda_1$ with $0<\lambda_1<\lambda$.

I've been trying with simple one variable functions but it turns out it always work. I guess it has to do with continuity. Is it always true? If it is, how to prove it?

An example I tried:

$(x+\lambda\cdot 1)^2\le x^2 + 1\cdot 1\cdot 2x\implies (x+\lambda^2)<x^2 + 2x$, solving for $\lambda$ gives an interval

Answer 1

I think the answer is no. Let $n=1,$ $f(t)= \sin^2 t.$ Set $x=0, \lambda = \pi, p=1,c=1/2.$ Note that $f'(0)=0.$ So we have

$$f(x+\lambda p) = f(\pi) = 0\, \le f(0) + (1/2)\pi f'(0) =0.$$

But for any $\lambda_1, 0 < \lambda_1 < \lambda,$ and any $c_1\in (0,1),$

$$f(x+\lambda_1p) = \sin^2( \lambda_1) >0,$$

while $f(x) + c_1\lambda_1f'(x) =0.$

Answer 2

A counterexample: take $f(x)=-(1+x^{10})\sin(\pi x)$ at $x=0$, $p=1$.

On short range close to $x=0$ it behaves as $-\sin(\pi x)$. However, further away the oscillations become more and more extreme.

For example, $\lambda=2.7$ will satisfy the Armijo's rule for any $c\in(0,1)$, but smaller $\lambda_1\in (1,2)$ will give positive functional values and, hence, not satisfying the rule again for any $c\in(0,1)$.