[proofplan] The local equivalence between $\Phi$ and $\|\cdot\|_X^2$ turns small initial norm into a small Lyapunov value, and monotonicity of $\Phi$ then keeps the whole trajectory in any prescribed small ball. For convergence, we take an initial datum in a sufficiently small sublevel set and study its omega-limit set. Precompactness gives a nonempty compact omega-limit set, while monotonicity of $\Phi$ forces $\Phi$ to be constant along complete trajectories inside that limit set. The hypothesis on complete constant-$\Phi$ trajectories then identifies the omega-limit set with $\{0\}$, and precompactness upgrades this uniqueness of omega-limit points to convergence of the whole positive orbit. [/proofplan]

[step:Choose a Lyapunov sublevel neighborhood inside the region of validity] Define \begin{align*} \alpha_0:=\min\{\rho,c_1r^2\}. \end{align*} For every $x\in X$ with $\Phi(x)<\alpha_0$, we have $x\in\mathcal U_\rho$, hence the trajectory $t\mapsto S(t)x$ remains in $B_X(0,r)$ and $t\mapsto \Phi(S(t)x)$ is non-increasing. Therefore, for every $t\geq 0$, \begin{align*} \Phi(S(t)x)\leq \Phi(x)<\alpha_0. \end{align*} Since $S(t)x\in B_X(0,r)$, the lower quadratic bound applies at $S(t)x$ and gives \begin{align*} c_1\|S(t)x\|_X^2\leq \Phi(S(t)x)<\alpha_0. \end{align*} Thus every trajectory starting in the sublevel set $\{\Phi<\alpha_0\}$ remains in $B_X(0,r)$ and satisfies \begin{align*} \|S(t)x\|_X<\sqrt{\frac{\alpha_0}{c_1}}\leq r \end{align*} for all $t\geq 0$. [/step]

[step:Prove Lyapunov stability from the quadratic bounds]Let $\varepsilon>0$ be given. Define \begin{align*} \alpha_\varepsilon:=\min\{\rho,c_1r^2,c_1\varepsilon^2\}. \end{align*} Choose \begin{align*} \delta_\varepsilon:=\min\left\{r,\sqrt{\frac{\alpha_\varepsilon}{c_2}}\right\}. \end{align*} If $\|x\|_X<\delta_\varepsilon$, then $\|x\|_X<r$, so the upper quadratic bound gives \begin{align*} \Phi(x)\leq c_2\|x\|_X^2<\alpha_\varepsilon\leq \rho. \end{align*} Hence $x\in\mathcal U_\rho$, and the trajectory remains in $B_X(0,r)$ and satisfies $\Phi(S(t)x)\leq\Phi(x)<\alpha_\varepsilon$ for all $t\geq0$. Applying the lower quadratic bound at $S(t)x\in B_X(0,r)$ gives \begin{align*} c_1\|S(t)x\|_X^2\leq \Phi(S(t)x)<\alpha_\varepsilon\leq c_1\varepsilon^2. \end{align*} Therefore $\|S(t)x\|_X<\varepsilon$ for every $t\geq0$. This proves Lyapunov stability of the equilibrium $0$.[/step]

[guided]We need to show that smallness of the initial norm forces smallness of the whole future orbit. The reason the functional $\Phi$ is useful is that it is comparable to the squared norm near $0$, and it decreases along the trajectory. Fix a target radius $\varepsilon>0$. We define the allowed Lyapunov level by \begin{align*} \alpha_\varepsilon:=\min\{\rho,c_1r^2,c_1\varepsilon^2\}. \end{align*} The three terms have different roles. The inequality $\alpha_\varepsilon\leq\rho$ ensures that initial data with $\Phi(x)<\alpha_\varepsilon$ lie in the sublevel set where the semiflow is global and $\Phi$ is non-increasing. The inequality $\alpha_\varepsilon\leq c_1r^2$ is compatible with the ball where the lower quadratic estimate is valid. The inequality $\alpha_\varepsilon\leq c_1\varepsilon^2$ is what will give the final estimate $\|S(t)x\|_X<\varepsilon$. Now choose \begin{align*} \delta_\varepsilon:=\min\left\{r,\sqrt{\frac{\alpha_\varepsilon}{c_2}}\right\}. \end{align*} If $\|x\|_X<\delta_\varepsilon$, then $x\in B_X(0,r)$, so the upper quadratic bound is available at $x$. It gives \begin{align*} \Phi(x)\leq c_2\|x\|_X^2<c_2\delta_\varepsilon^2\leq \alpha_\varepsilon. \end{align*} Since $\alpha_\varepsilon\leq\rho$, this puts $x$ in $\mathcal U_\rho$. By hypothesis, the trajectory $t\mapsto S(t)x$ remains in $B_X(0,r)$ and the function $t\mapsto \Phi(S(t)x)$ is non-increasing. Hence, for every $t\geq0$, \begin{align*} \Phi(S(t)x)\leq \Phi(x)<\alpha_\varepsilon. \end{align*} Because $S(t)x\in B_X(0,r)$, the lower quadratic bound applies at the point $S(t)x$: \begin{align*} c_1\|S(t)x\|_X^2\leq \Phi(S(t)x)<\alpha_\varepsilon\leq c_1\varepsilon^2. \end{align*} Dividing by $c_1>0$ and taking square roots gives $\|S(t)x\|_X<\varepsilon$ for every $t\geq0$. This is exactly Lyapunov stability of the equilibrium $0$.[/guided]

[step:Construct omega-limit trajectories on which the Lyapunov functional is constant]Choose \begin{align*} \delta_0:=\min\left\{r,\sqrt{\frac{\alpha_0}{c_2}}\right\}. \end{align*} Let $x\in X$ satisfy $\|x\|_X<\delta_0$. As in the previous step, $\Phi(x)<\alpha_0\leq\rho$, so $x\in\mathcal U_\rho$. Define the positive orbit \begin{align*} \mathcal O_x:=\{S(t)x:t\geq0\} \end{align*} and the omega-limit set \begin{align*} \omega(x):=\{y\in X:\text{there exists a sequence }(t_k)_{k=1}^{\infty}\subset[0,\infty)\text{ with }t_k\to\infty\text{ and }S(t_k)x\to y\text{ in }X\}. \end{align*} By hypothesis, $\mathcal O_x$ is precompact in $X$. Hence $\omega(x)$ is nonempty: for any sequence $t_k\to\infty$, the sequence $(S(t_k)x)_{k=1}^{\infty}$ has a convergent subsequence in $X$. It is also contained in $\overline{\mathcal O_x}\subset \overline{B_X(0,r)}$. The function $\varphi_x:[0,\infty)\to[0,\infty)$ defined by $\varphi_x(t):=\Phi(S(t)x)$ is non-increasing and bounded below by $0$, so the limit \begin{align*} \ell:=\lim_{t\to\infty}\Phi(S(t)x) \end{align*} exists in $[0,\infty)$. If $y\in\omega(x)$ and $t_k\to\infty$ is a sequence with $S(t_k)x\to y$, then continuity of $\Phi$ gives \begin{align*} \Phi(y)=\lim_{k\to\infty}\Phi(S(t_k)x)=\ell. \end{align*} Thus $\Phi$ is equal to the single value $\ell$ on $\omega(x)$. We next record the invariance property needed below. Let $y\in\omega(x)$. For every $\tau\geq0$, continuity of $S(\tau):X\to X$ and the semiflow identity give \begin{align*} S(\tau)y=\lim_{k\to\infty}S(\tau)S(t_k)x=\lim_{k\to\infty}S(t_k+\tau)x. \end{align*} Since $t_k+\tau\to\infty$, this proves $S(\tau)y\in\omega(x)$. Hence every positive trajectory starting in $\omega(x)$ stays in $\omega(x)$, and $\Phi(S(\tau)y)=\ell$ for every $\tau\geq0$. It remains to pass from positive invariance to complete trajectories. Fix $y\in\omega(x)$, and choose a sequence $(t_k)_{k=1}^{\infty}\subset[0,\infty)$ with $t_k\to\infty$ and $S(t_k)x\to y$. After discarding finitely many terms for each fixed integer $m\geq1$, the sequence $S(t_k-m)x$ is defined and belongs to $\mathcal O_x$. Since $\mathcal O_x$ is precompact, a diagonal subsequence, not relabeled, may be chosen so that for every $m\geq1$ there is a point $y_{-m}\in X$ with \begin{align*} S(t_k-m)x\to y_{-m} \end{align*} in $X$ as $k\to\infty$. Because $t_k-m\to\infty$, each $y_{-m}$ belongs to $\omega(x)$. Continuity of $S(m):X\to X$ gives \begin{align*} S(m)y_{-m}=\lim_{k\to\infty}S(m)S(t_k-m)x=\lim_{k\to\infty}S(t_k)x=y. \end{align*} For consistency, fix $m\geq1$. Passing to the limit in the identity $S(1)S(t_k-(m+1))x=S(t_k-m)x$ and using continuity of $S(1):X\to X$ gives \begin{align*} S(1)y_{-(m+1)}=y_{-m}. \end{align*} Moreover, since $\Phi(y_{-m})=\ell<\rho$, each $y_{-m}$ lies in $\mathcal U_\rho\subset B_X(0,r)$. Define \begin{align*} u_y:\mathbb R\to X \end{align*} by setting $u_y(t):=S(t+m)y_{-m}$ whenever $m\in\mathbb N$ and $t\in[-m,-m+1]$, and by setting $u_y(t):=S(t)y$ for $t\geq0$. The consistency relation makes this definition independent of the interval endpoint. The semiflow identity gives \begin{align*} u_y(t+s)=S(t)u_y(s) \end{align*} for every $s\in\mathbb R$ and every $t\geq0$. Since $\ell\leq\Phi(x)<\alpha_0\leq\rho$, we have $\Phi(u_y(t))=\ell<\rho$ for every $t\in\mathbb R$, hence $u_y(t)\in\mathcal U_\rho\subset B_X(0,r)$ for every $t\in\mathbb R$. Thus $u_y$ is a complete trajectory in $B_X(0,r)$ on which $\Phi$ is constant.[/step]

[guided]The purpose of this step is to extract the part of LaSalle's principle that is needed here, directly from precompactness and continuity of the semiflow. First define the positive orbit \begin{align*} \mathcal O_x:=\{S(t)x:t\geq0\} \end{align*} and the omega-limit set \begin{align*} \omega(x):=\{y\in X:\text{there exists a sequence }(t_k)_{k=1}^{\infty}\subset[0,\infty)\text{ with }t_k\to\infty\text{ and }S(t_k)x\to y\text{ in }X\}. \end{align*} Because $x\in\mathcal U_\rho$, the orbit is global and remains in $B_X(0,r)$. By the precompactness hypothesis, the closure of $\mathcal O_x$ is compact in $X$. Therefore every sequence of orbit points with times tending to infinity has a convergent subsequence, so $\omega(x)$ is nonempty. The Lyapunov functional has a limit along the orbit. Indeed, the function $\varphi_x:[0,\infty)\to[0,\infty)$ defined by $\varphi_x(t):=\Phi(S(t)x)$ is non-increasing by hypothesis and is bounded below by $0$. Hence the monotone convergence property for real-valued functions gives a limit \begin{align*} \ell:=\lim_{t\to\infty}\Phi(S(t)x). \end{align*} Now take any $y\in\omega(x)$. By definition, there is a sequence $t_k\to\infty$ such that $S(t_k)x\to y$ in $X$. Since $\Phi:X\to[0,\infty)$ is continuous, \begin{align*} \Phi(y)=\lim_{k\to\infty}\Phi(S(t_k)x)=\ell. \end{align*} Thus every omega-[limit point](/page/Limit%20Point) lies on the same Lyapunov level $\{\Phi=\ell\}$. Next we show that omega-limit points generate forward orbits that remain inside the omega-limit set. Fix $y\in\omega(x)$ and $\tau\geq0$. Choose $t_k\to\infty$ with $S(t_k)x\to y$. The map $S(\tau):X\to X$ is continuous, and the semiflow identity gives \begin{align*} S(\tau)S(t_k)x=S(t_k+\tau)x. \end{align*} Passing to the limit, \begin{align*} S(\tau)y=\lim_{k\to\infty}S(t_k+\tau)x. \end{align*} Since $t_k+\tau\to\infty$, this proves $S(\tau)y\in\omega(x)$. Consequently $\Phi(S(\tau)y)=\ell$ for every $\tau\geq0$. The hypothesis of the theorem is stated for complete trajectories, so we must also obtain a backward extension through each $y\in\omega(x)$. Fix $y\in\omega(x)$, and choose one sequence $(t_k)_{k=1}^{\infty}\subset[0,\infty)$ with $t_k\to\infty$ and $S(t_k)x\to y$. Using a single sequence is important: separate choices for different backward times would not automatically give compatible preimages. For each fixed integer $m\geq1$, all sufficiently large $k$ satisfy $t_k\geq m$, so the shifted points $S(t_k-m)x$ are defined and belong to the positive orbit. Since the orbit is precompact in $X$, we can perform the usual diagonal extraction: first choose a subsequence along which $S(t_k-1)x$ converges, then a further subsequence along which $S(t_k-2)x$ converges, and so on; taking the diagonal subsequence gives one subsequence, still denoted $(t_k)$, such that for every fixed $m\geq1$ there exists $y_{-m}\in X$ with \begin{align*} S(t_k-m)x\to y_{-m} \end{align*} in $X$. Since $t_k-m\to\infty$, the definition of $\omega(x)$ gives $y_{-m}\in\omega(x)$. Now the desired compatibility follows from the semiflow identity and continuity. For each $m\geq1$, \begin{align*} S(m)y_{-m}=\lim_{k\to\infty}S(m)S(t_k-m)x=\lim_{k\to\infty}S(t_k)x=y. \end{align*} Likewise, passing to the limit in $S(1)S(t_k-(m+1))x=S(t_k-m)x$ gives \begin{align*} S(1)y_{-(m+1)}=y_{-m}. \end{align*} Thus the backward points are consistent. Also, because each $y_{-m}\in\omega(x)$ and $\Phi$ equals $\ell$ on $\omega(x)$, we have $\Phi(y_{-m})=\ell<\rho$; hence $y_{-m}\in\mathcal U_\rho\subset B_X(0,r)$. Define \begin{align*} u_y:\mathbb R\to X \end{align*} by $u_y(t):=S(t)y$ for $t\geq0$, and by $u_y(t):=S(t+m)y_{-m}$ when $m\in\mathbb N$ and $t\in[-m,-m+1]$. The consistency relation ensures that the definitions agree at the integer endpoints. The semiflow identity then gives \begin{align*} u_y(t+s)=S(t)u_y(s) \end{align*} for every $s\in\mathbb R$ and every $t\geq0$. Since all points used in the construction lie in $\omega(x)$ and $\omega(x)$ is positively invariant, $u_y(t)\in\omega(x)$ for every $t\in\mathbb R$. Therefore $\Phi(u_y(t))=\ell$ for every $t\in\mathbb R$. We have constructed a complete trajectory in $B_X(0,r)$ on which $\Phi$ is constant.[/guided]

[step:Identify the omega-limit set with the zero equilibrium] Let $y\in\omega(x)$. The previous step constructs a complete trajectory \begin{align*} u_y:\mathbb R\to B_X(0,r) \end{align*} with $u_y(0)=y$ and with $\Phi(u_y(t))$ constant for all $t\in\mathbb R$. By the hypothesis on complete constant-$\Phi$ trajectories, $u_y(t)=0$ for every $t\in\mathbb R$. In particular $y=u_y(0)=0$. Since $y\in\omega(x)$ was arbitrary and $\omega(x)$ is nonempty, we conclude \begin{align*} \omega(x)=\{0\}. \end{align*} [/step]

[step:Upgrade the singleton omega-limit set to convergence of the full orbit] We prove that $\|S(t)x\|_X\to0$. Suppose not. Then there exist $\varepsilon_*>0$ and a sequence $(t_k)_{k=1}^{\infty}\subset[0,\infty)$ with $t_k\to\infty$ such that \begin{align*} \|S(t_k)x\|_X\geq\varepsilon_* \end{align*} for every $k\in\mathbb N$. Since the orbit $\mathcal O_x$ is precompact, after passing to a subsequence there exists $z\in X$ such that \begin{align*} S(t_k)x\to z \end{align*} in $X$. Because $t_k\to\infty$, the definition of $\omega(x)$ gives $z\in\omega(x)$. From the previous step, $\omega(x)=\{0\}$, so $z=0$. This contradicts $\|S(t_k)x\|_X\geq\varepsilon_*$ for every $k$, since convergence in $X$ to $0$ implies $\|S(t_k)x\|_X\to0$ along the subsequence. Hence \begin{align*} \lim_{t\to\infty}\|S(t)x\|_X=0. \end{align*} The choice of $\delta_0$ above shows that every initial datum satisfying $\|x\|_X<\delta_0$ has a positive trajectory converging to $0$, while the earlier Lyapunov estimate proves stability. Therefore $0$ is an asymptotically stable equilibrium. [/step]