I'm reading below theorem from this lecture note.
Theorem 4.5. Let $H$ be a Hilbert space over $\mathbb{R}$. If $\varphi: H \rightarrow \mathbb{R}$ is differentiable and convex, then for every $u \in H$ there exists a unique $y:[0, \infty) \rightarrow H$ such that $$ \begin{aligned} & y^{\prime}(t)=-\nabla \varphi(y(t)), \quad t \geq 0, \\ & y(0)=u. \end{aligned} $$
Idea of proof. Fix $h>0$. Discretize the differential equation with Euler's implicit scheme, $$ \frac{y((n+1) h)-y(n h)}{h}=-\nabla \varphi(\color{red}{y(n h)}) . $$ Then, with $y_n=y(n h)$, $$ y_{n+1}+h \nabla \varphi\left(\color{red}{y_{n+1}}\right)=y_n, $$ so that $$ y_{n+1}=(I+h \nabla \varphi)^{-1}\left(y_n\right) . $$ One of the difficulties is to show that the function $x \mapsto x+h \nabla \varphi(x)$ is invertible. Then $$ J_h(x):=(I+h \nabla \varphi)^{-1}(x), \quad x \in H, $$ is called the resolvent associated to $\nabla \varphi$. We obtain $$ y_n=J_h^n(u), $$ and this is hoped to be a good approximation for $y(n h)$. The next steps are to show that $$ y(t):=\lim _{k \rightarrow \infty} J_{t / k}^n(u) $$ exists and that the function $y$ thus defined is the unique solution.
My understanding It seems to me $$ \frac{y((n+1) h)-y(n h)}{h}=-\nabla \varphi(\color{red}{y(n h)}) $$ implies $y_{n+1}+h \nabla \varphi\left(\color{red}{y_{n}}\right)=y_n$, and not $y_{n+1}+h \nabla \varphi\left(\color{red}{y_{n+1}}\right)=y_n$ as written in the lecture note. Then it should be $y_{n}=(I-h \nabla \varphi)^{-1}\left(y_{n+1}\right)$.
Could you confirm if my understanding is correct, or I miss something else?
Since the author mentioned Euler's implicit scheme (i.e., Backward Euler's method), the approximation should be, with $y_n\colon = y(nh)$, $$ \frac{y_{n + 1} - y_n}{h} = -\nabla\varphi(y_{n + 1}). $$