In the following, I am referring to this paper, p. 14, line (3.14):

It is said that from $$ \frac{\Vert f^{n+1}\Vert_{2,\gamma}^2-\Vert f^n\Vert_{2,\gamma}^2}{2\Delta t}+\frac{1}{\varepsilon^2}\Vert f^{n+1}-\rho^{n+1}\mathcal{M}\Vert_{2,\gamma}^2\leq 0\tag{3.13} $$ for all $n\geq 0$ it particularly follows that $$ \max\left(\sup_{n\geq 0}\Vert f^n\Vert_{2,\gamma}^2,\frac{2}{\varepsilon^2}\sum_{n=1}^\infty\Delta t\Vert f^n-\rho^n\mathcal{M}\Vert_{2,\gamma}^2\right)\leq\Vert f^0\Vert_{2,\gamma}^2\tag{3.14} $$

I have two questions:

(1) Why does (3.13) imply (3.14)?

(2) It is said that (3.14) implies uniqueness of a solution with respect to (3.10) because (3.10) is a finite dimensional linear system. Why does this imply the uniqueness of a solution? As a finite dimensional linear system, it has either no solution, a unique solution or infinitely many solutions. So how does (3.14) tell me that there exists a unique solution?

As to (1):

(3.13) is equivalent to $$ \Vert f^{n+1}\Vert_{2,\gamma}^2+\frac{2}{\varepsilon^2}\Delta t \Vert f^{n+1}-\rho^{n+1}\mathcal{M}\Vert_{2,\gamma}^2\leq \Vert f^n\Vert_{2,\gamma}^2 $$

Now, as this holds for all $n\geq 0$, I can take the supremum over $n\geq 0$ on both sides. But what next?


I have outsourced question (2) to a separate question!


1 Answer 1


No idea about your second question since it’s too much notation to parse through but (3.13) is an inequality of the form \begin{align} \frac{a_{n+1}-a_n}{2\Delta t}+b_{n+1}\leq 0, \end{align} where each $a_n,b_n\geq 0$. Now, multiply by $2\Delta t$, and given an integer $N\geq 1$, sum from $n=0$ to $N-1$, and simplify the telescoping sum to get \begin{align} a_{N}-a_0 +2\Delta t\sum_{n=0}^{N-1}b_{n+1}&\leq 0, \end{align} or by re-indexing, and rearranging, you get \begin{align} a_{N}+2\Delta t\sum_{n=1}^{N}b_n&\leq a_0. \end{align} By non-negativity, this tells us that for all $N\geq 1$, we have $a_N\leq a_0$ and that for all $N\geq 1$, $2\Delta t\sum_{n=1}^Nb_n\leq a_0$. Now, take the supremum over all $N\geq 1$ (and note that the supremum of a finite sum of non-negative numbers is equal to the series) to get \begin{align} \sup_{N\geq 1}a_N & \leq a_0 \quad\text{and}\quad \sup_{N\geq 1}2\Delta t\sum_{n=1}^Nb_n=2\Delta t\sum_{n=1}^{\infty}b_n\leq a_0. \end{align} In addition to the first inequality, we obviously have $a_0\leq a_0$ as well, so we can take the supremum over all $N\geq 0$. Having these two inequalities implies the maximum is also bounded by $a_0$, hence \begin{align} \max\left(\sup_{n\geq 0}a_n,2\Delta t\sum_{n=1}^{\infty}b_n\right)&\leq a_0. \end{align}

  • $\begingroup$ Thanks a lot, great explanation! $\endgroup$
    – selector
    Jun 5 at 10:11

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