Suppose I have nonlinear operator $T: V \to V$ where $V$ is a finite dimensional Hilbert space with inner product $( \cdot, \cdot )$. I need to prove that $T$ is strongly monotone, i.e. there exists $m > 0$ such that $( T(u) - T(v), u - v) \geq m \|u - v\|^2$ for all $u,v \in V$. Moreover, $T = I + \tau D + \tau f$, where $I$ is the identity operator, $D$ is positive definite and linear, and $f$ is nonlinear but is Lipschitz continuous. $\tau$ represents a time step, and can be controlled to be as small as necessary.
It's easy to see that
\begin{align*} (T(u) - T(v), u-v) &= ((I + \tau D)(u-v) + \tau (f(u) - f(v)), u-v) \\ &= ((I + \tau D)(u - v), u - v) + \tau(f(u) - f(v), u - v), \end{align*}
and that $((I + \tau D)(u-v), u-v) > \|u - v\|^2$ because the eigenvalues of $I + \tau D$ are all strictly greater than 1.
I suspect that the idea from here is to use that $f$ is Lipschitz to argue that there is some small-enough $\tau$ so that the contribution of $\tau((f(u) - f(v),u-v)$ is small enough. Specifically, I was thinking that I could use Cauchy-Schwartz to show that
\begin{align} (f(u)-f(v), u-v) \leq \|f(u) - f(v)\| \|u - v\| \leq M \|u - v\|^2. \end{align}
And then, somehow, the fact that $(f(u) - f(v), u -v)$ is bounded above for all $u,v$ means that it doesn't throw off the whole calculation too much. But....it just...somehow the pieces don't all fit together.
Any ideas?
Nevermind I figured it out. $(f(u) - f(v), u - v)$ could be positive or negative. Since we seek a lower bound, assume it is negative and write
\begin{align*} (T(u) - T(v), u - v) = ((I + \tau D)(u-v), u - v) - \tau|(f(u) - f(v), u - v)|. \end{align*}
Then, use Cauchy Schwartz, the Lipschitz condition, and the fact that $\tau$ may be made arbitrarily small, as indicated in my original post.
(I swear, if they end up giving me a Ph.D., my brain-dead self will be as surprised as anyone.)