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In this paper, the stability of a discretization $$L_hx=y~~~~~~~~~~~(*) $$on a continuous problem $$Lx=y~~~~~~~~~~~(**)$$ where $L\in BL(X,Y)$ and $h$ represents the mesh parameters, is given by the uniform boundedness of the inverse operator $L_h^{-1}$. That is, the numerical scheme $(*)$ is said to be stable if $$\lVert L_h^{-1} \lVert \leq C,$$ for some $C>0$ irrespective of the family of mesh parameters represented by $h$.

My doubts:

How can we ensure the boundedness of the most often partial differential operators appear in $(*)$ as we have seen that derivative operators are not continuous?

How can we practically explain the connection between stability and the 'uniform boundedness'?

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    $\begingroup$ Most derivative operators are continuous in the appropriate norms (e.g. $\frac{\partial}{\partial x}\in W^{1,1}\to L^1$); see Sobolev spaces. As for why this is a useful definition, see Thm. 1.1 in the paper you cite. $\endgroup$
    – Jacob Manaker
    Jun 23 at 9:54

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One notion of stability is that "for small differences in input, we get small differences in output". Mathematically, it could be, e.g., for two pieces of data $y_1, y_2$ and their respective solutions $x_1,x_2$, you would want $$\|x_1-x_2\| \leq C \|y_1-y_2\|.$$

If $y_1$ and $y_2$ are 'close' in some sense, then the solutions to the inverse problem should also be 'close', perhaps up to an extra multiplicative factor $C$. The uniform boundedness of $L_h^{-1}$ gives precisely this notion of stability, since (for this system) a solution is produced via $x=L_h^{-1}y$. So

$$\|x_1-x_2\|=\|L_h^{-1}y_1-L_h^{-1}y_2\|\leq\|L_h^{-1}\|\|y_1-y_2\|\leq C\|y_1-y_2\|.$$

Why is $L_h$ bounded?: Since (as far as I can tell) $L_h$ is acting on a finite dimensional discretization. Even though it represents differentiation, it is still bounded since it is acting on a finite dimensional space (and all linear operators in finite dimensional spaces are bounded).

Concerning $(**)$: this is a very good observation, and in general, a linear operator involving differentiation will not be bounded in an infinite-dimensional space (and its inverse may not even exist)! So very careful analysis must be done in order to ensure that, as the grid $h$ refines, you actually converge to a solution of the continuous problem. Check out the literature/proofs on Galerkin methods for more info on this :-)

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    $\begingroup$ Sir, Which is that finite dimensional vector space domain of $L_h$? $\endgroup$
    – Messi Lio
    2 days ago
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    $\begingroup$ @MessiLio The discrete spaces mentioned at the start of section 2 of the paper in your post $\endgroup$
    – Zim
    2 days ago
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    $\begingroup$ Sir, I couldn't understand how the discrete space $X_h$ is a finite dimensional subspace of $X$ ('in some cases'). Can you make it clear ? $\endgroup$
    – Messi Lio
    2 days ago
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    $\begingroup$ @MessiLio ok a very simple example is as follows: Consider all continuous functions on $[0,1]$. This is an infinite-dimensional space. However, if we fixed a particular grid, say $h=\{0,0.5,1\}$, then a "function" defined on $X_h$ is just a mapping from each point in $h$ to $\mathbb{R}$, i.e., we represent a function $f$ with a vector in (the finite dimensional space) $\mathbb{R}^3$ which represents $(f(0), f(0.5), f(1))$. So, in general in 1D, the cardinality of your discretization is the dimension of your space. $\endgroup$
    – Zim
    2 days ago
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    $\begingroup$ @MessiLio a lot of folks use a slight abuse of notation here. For any function in $X$, you can construct is representation in $X_h$ via just taking its values on the grid. For any real numbers representing the values of $f$ at the gridpoints, you can construct a representative function in $X$ in the following procedure: For every grid point, define a "hat function" on $X$ which takes the value of $1$ at that gridpoint, and $0$ at other gridpoints.The linear combination of the function values and the hat functions is a 'representation' of the finite dimensional "function" in $X$. $\endgroup$
    – Zim
    2 days ago

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