Brezis' lemma 7.1: is it true that $w$ is $C^1$ implies $w$ is $C^\infty$?

Question

Let $H$ together with an inner product $\langle \cdot, \cdot \rangle$ be a real Hilbert space. Let $|\cdot|$ be the induced norm. Let $A_\lambda :H \to H$ be a bounded linear operator. I'm reading Lemma 7.1 in Brezis' Functional Analysis, i.e.,

Let $w \in C^1 ([0,+\infty) ; H)$ be a function satisfying $$ \frac{d w}{d t}+A_\lambda w=0 \quad \text {on} \quad [0,+\infty). $$ Then the functions $t \mapsto|w(t)|$ and $t \mapsto\left|\frac{d w}{d t}(t)\right|=\left|A_\lambda w(t)\right|$ are non-increasing on $[0,+\infty)$.

My understanding

1. By assumption, $w$ belongs to $C^1 ([0,+\infty) ; H)$. Because $A_\lambda$ is bounded linear, it belongs to $C^\infty (H ; H)$. By chain rule, $A_\lambda w$ belongs to $C^1 ([0,+\infty) ; H)$. Because $\frac{d w}{d t} = -A_\lambda w$, we get $\frac{d w}{d t}$ belongs to $C^1 ([0,+\infty) ; H)$. Then $w$ belongs to $C^2 ([0,+\infty) ; H)$. Repeating this argument, we get $w$ belongs to $C^k ([0,+\infty) ; H)$ for any $k \in \mathbb N$. So $w$ belongs to $C^\infty ([0,+\infty) ; H)$.

2. It seems above reasoning still holds (and thus $w \in C^\infty ([0,+\infty) ; H)$) even if we relax $A_\lambda$ to be just continuously differentiable.

Could you please confirm if my understanding is fine?

Answer 1

As pointed out by @PhoemueX, (2) is not correct. Luckily, (1) is correct. Indeed, the author mentions in his proof that

On the other hand, since $A_\lambda$ is a linear bounded operator, we deduce (by induction) that $w \in C^{\infty}([0,+\infty) ; H)$ and also that $$ \frac{d}{d t}\left(\frac{d w}{d t}\right)+A_\lambda\left(\frac{d w}{d t}\right)=0. $$