I am self-studying the following material, whose source is https://sayanmuk.github.io/StochasticAnalysisManifolds.pdf
I am stuck at the final step of the following propsotion proof.
Proposition 1.2.8. Let $X$ be the solution of the extended equation (1.2.3) up to its explosion time $e(X)$ and $X_0 \in M$. Then $X_t \in M$ for $0 \leq t < e(X)$.
Proof. Without loss of generality, we may assume that $X_0$ is uniformly bounded (see the end of the proof of Theorem 1.1.8). Let $d_{RN}(x, M)$ be the Euclidean distance from $x$ to $M$. Because $M$ is closed and without boundary, the function $f(x) = d_{RN}(x, M)^2$ is smooth in a neighborhood of $M$. Multiplying by a suitable cut-off function, we may assume that $f \in C^{\infty}(\mathbb{R}^N)$.
Since the vector fields $V_{\alpha}$ are tangent to $M$ along the submanifold $M$, a local calculation shows that the functions $V_{\alpha}f$ and $V_{\alpha}V_{\beta}f$ vanish along $M$ at a rate proportional to the square of the distance $d_{RN}(x, M)$. Hence, there exists a neighborhood $U$ of $M$ such that:
(i) $f(x) = 0$ for $x \in U$ if and only if $x \in M$;
(ii) for each $R > 0$, there exists $C$ depending on $R$ such that $|V_{\alpha}f(x)| \leq Cf(x)$, $|V_{\alpha}V_{\beta}f(x)| \leq Cf(x)$ for all $x \in U \cap B(0, R)$.
Define the stopping times: $$ \tau_R = \inf\{ t > 0 : X_t \notin B(R) \}, $$ $$ \tau_U = \inf\{ t > 0 : X_t \notin U \}, $$ $$ \tau = \tau_U \wedge \tau_R. $$
Consider the process $X$ before $\tau$, the first exit time from $U \cap B(R)$. Applying Itô's formula, we have: $$ f(X_t) = \int_0^t V_{\alpha}f(X_s) dZ_s^{\alpha} + \frac{1}{2} \int_0^t V_{\alpha}V_{\beta}f(X_s) d\langle Z^{\alpha}, Z^{\beta} \rangle_s, $$ where $Z^{\beta}$ is.
Using Lemma 1.1.2 to the right side of the above equation, we have: $$ \mathbb{E}|f(X)|^2_{\infty, \eta t \wedge \tau} \leq C \sum_{\alpha, \beta} \int_0^{\eta t \wedge \tau} (|V_{\alpha}f(X_s)|^2 + |V_{\alpha}V_{\beta}f(X_s)|^2) ds. $$
We now make the time change $s \mapsto \eta s$ and use (1.2.4), which is permissible because $X_s \in U \cap B(R)$ if $s \leq \tau$. This yields the inequality: $$ \mathbb{E}|f(X)|^2_{\infty, \eta t \wedge \tau} \leq C_1 \int_0^t \mathbb{E}|f(X)|^2_{\infty, \eta s \wedge \tau} ds. $$
It follows that $f(X_s) = 0$ for $s \leq \eta t \wedge \tau$, or $X_s \in M$ for $0 \leq s < \eta t \wedge \tau_U \wedge \tau_R$.
Letting $t$ and then $R$ go to infinity, we see that $X_s \in M$ for $0 \leq s < e(X) \wedge \tau_U$, from which we conclude easily that $X$ stays on $M$ up to $e(X)$. $\blacksquare$
Question How can the author know that f(x)=0 or $X_s\in M$ by looking at $$ \mathbb{E}|f(X)|^2_{\infty, \eta t \wedge \tau} \leq C_1 \int_0^t \mathbb{E}|f(X)|^2_{\infty, \eta s \wedge \tau} ds. $$?
$C_1$ can be bigger than 1 and so, it does not require $f(x)=0$, at least I do not see how.
Can someone explain me this step?
Thanks in advance!
Edit: Here is the definition of explosion time:
Definition 1.1.4.: An $M$-valued path $x$ with explosion time $e = e(x) > 0$ is a continuous map $x : [0,\infty) \to M_c$ such that $x_t \in M$ for $0 \leq t < e$ and $x_t = \partial M$ for all $t \geq e$ if $e < \infty$. The space of $M$-valued paths with explosion time is called the path space of $M$ and is denoted by $W(M)$.