Well-Posedness: Coercivity, Inf-Sup, and Why We Bother
In the previous post, we wrote parametric PDEs as an abstract operator equation $B_\mu u_\mu = f_\mu$. That’s a clean way to state a problem — but stating a problem isn’t the same as knowing it has a solution. This post is about why we reformulate PDEs the way we do, and what machinery that reformulation unlocks.
The trouble with the classical form
Classically, a second-order PDE demands that its solution $u$ be twice continuously differentiable, $u \in C^2$. That’s a strong requirement, and physical reality routinely breaks it:
- Heterogeneous materials — a sudden jump in diffusivity where metal meets wood (exactly the thermal block from the last post) means $u$ generically isn’t twice differentiable right at the interface.
- Non-convex domains — an L-shaped room has a re-entrant corner where classical solutions can develop singularities.
- Point sources — a right-hand side $f_\mu$ concentrated at a single point isn’t even a function in the classical sense.
In any of these cases, a $C^2$ solution simply doesn’t exist. If it doesn’t exist, the problem is ill-posed — and yet these are exactly the situations we want to model.
The fix: weaken the requirements
Multiply the PDE by a test function $v$ and integrate over the domain. Then integrate by parts (Green’s theorem) to move one derivative off of $u$ and onto $v$:
- Before: $u$ had to be differentiated twice.
- After: $u$ only needs one (weak) derivative, and so does $v$.
This is exactly why Sobolev spaces like $H^1$ showed up in the last post — they only demand first weak derivatives and finite energy. By widening the space we search in, a solution can exist again, even for jump coefficients, corners, and point sources.
Moving into these Hilbert spaces isn’t just a technical convenience — it unlocks theorems that automatically certify well-posedness (existence, uniqueness, stability), which is what the rest of this post is about.
Coercivity: the variational analogue of ellipticity
Recall ellipticity is a pointwise condition on the diffusion coefficient itself:
\[\xi^\top \boldsymbol\alpha_\mu(x)\, \xi \ge c\,|\xi|^2 \qquad \text{for all } \xi, \text{ at every } x.\]Coercivity is the corresponding condition on the bilinear form $b_\mu(u,v)$ that comes out of the variational formulation — a single, global statement over the whole domain rather than a pointwise one:
Definition — Coercivity
A bilinear form $b_\mu: \mathcal U \times \mathcal U \to \mathbb R$ is coercive (or $\mathcal U$-elliptic) if there is a constant $\alpha_\mu > 0$ such that
\(b_\mu(u,u) \;\ge\; \alpha_\mu \, \|u\|_{\mathcal U}^2 \qquad \text{for all } u \in \mathcal U.\)
In words: plugging the same function into both slots of the bilinear form never gives you something smaller than a fixed multiple of $|u|^2$. Nothing “cancels away to nothing” except when $u$ itself is zero.
The inf-sup condition
Coercivity requires $u$ and $v$ to live in the same space. Once trial and test spaces differ ($\mathcal U \ne \mathcal V$, a Petrov–Galerkin setup), coercivity doesn’t even make sense as a concept — you need something more general:
Definition — Inf-sup condition
$b_\mu$ satisfies the inf-sup condition if there is a constant $\beta_\mu > 0$ such that
\[\inf_{u \in \mathcal U}\; \sup_{v \in \mathcal V} \; \frac{b_\mu(u,v)}{\|u\|_{\mathcal U}\|v\|_{\mathcal V}} \;\ge\; \beta_\mu.\]Equivalently: for every $u \in \mathcal U$, there exists a $v \in \mathcal V$ with
\(\sup_{v \in \mathcal V} \frac{b_\mu(u,v)}{\|v\|_{\mathcal V}} \;\ge\; \beta_\mu\, \|u\|_{\mathcal U}.\)
Intuition: for every possible solution $u$, there’s at least one test function that “sees” it — if $u \ne 0$, the bilinear form doesn’t evaluate to zero against every $v$. No matter how you pick $u$, the test space is rich enough to detect it.
Proof: ellipticity implies coercivity
Proof — ellipticity of the diffusion coefficient implies coercivity of the bilinear form
Consider just the diffusion term of the bilinear form, with $v = u$:
\[b_\mu(u,u) = \int_\Omega \nabla u(x)^\top \boldsymbol\alpha_\mu(x) \nabla u(x)\, dx.\]Step 1 — apply ellipticity pointwise. Since $\boldsymbol\alpha_\mu(x)$ is positive definite with $\xi^\top\boldsymbol\alpha_\mu(x)\xi \ge c|\xi|^2$, plug in $\xi = \nabla u(x)$:
\[b_\mu(u,u) \;\ge\; \int_\Omega c\, |\nabla u(x)|^2\, dx \;=\; c\, |u|_{H^1}^2,\]where $|u|_{H^1}$ is the $H^1$ semi-norm — it only measures the derivative, not the function values themselves.
Step 2 — apply the Poincaré–Friedrichs inequality. Coercivity needs the full norm $|u|_{\mathcal U}^2$ on the right, not just the semi-norm. Because $u$ vanishes on the boundary (homogeneous Dirichlet BC), the Poincaré–Friedrichs inequality lets us bound the full $H^1$ norm by the derivative semi-norm alone:
\[c\, |u|_{H^1}^2 \;\ge\; \alpha_\mu \, \|u\|_{H^1}^2.\]Conclusion.
\[b_\mu(u,u) \;\ge\; \alpha_\mu\, \|u\|_{\mathcal U}^2.\]Ellipticity of the coefficient, plus Poincaré–Friedrichs (which needs the Dirichlet BC to hold), together give coercivity of the bilinear form.
Coercivity implies the inf-sup condition
Remark
If $b_\mu$ is coercive with constant $\alpha_\mu$, it automatically satisfies the inf-sup condition, with inf-sup constant $\beta_\mu = \alpha_\mu$.
Proof — coercivity implies inf-sup
To bound the supremum over all $v$, it’s enough to exhibit one good choice — so pick $v = u$:
\[\sup_{v \in \mathcal V} \frac{b_\mu(u,v)}{\|v\|} \;\ge\; \frac{b_\mu(u,u)}{\|u\|} \;\ge\; \frac{\alpha_\mu \|u\|^2}{\|u\|} \;=\; \alpha_\mu \|u\|.\]So the inf-sup constant is exactly the coercivity constant: $\beta_\mu = \alpha_\mu$.
This is a “get out of jail free” card: proving an inf-sup condition directly is often hard, but checking coercivity is easy (you just look for ellipticity) — and coercivity gives you inf-sup for free.
Lax–Milgram: the well-behaved case
Theorem — Lax–Milgram
Let $\mathcal U = \mathcal V$ be a Hilbert space, and suppose the bilinear form $a_\mu$ is
-
Continuous (bounded): $\lvert a_\mu(u,v) \rvert \le C_\mu |u|\,|v|$ for all $u,v$, and
-
Coercive: $a_\mu(u,u) \ge \alpha_\mu |u|^2$ for all $u$.
Then for every $f_\mu \in \mathcal V’$, there exists a unique $u_\mu \in \mathcal U$ with $a_\mu(u_\mu, v) = f_\mu(v)$ for all $v$, and $u_\mu$ depends continuously on $f_\mu$ (the problem is stable).
Diffusion-dominated problems — like the thermal block — are the classic Lax–Milgram case: trial and test spaces coincide, coercivity follows straightforwardly from ellipticity as shown above, and well-posedness comes essentially for free.
Banach–Nečas: the general case
Once advection dominates, or you’re dealing with the wave equation or the Schrödinger equation, coercivity often fails — and you may deliberately want $\mathcal U \ne \mathcal V$ (a Petrov–Galerkin setup, used heavily later for ultraweak formulations). Lax–Milgram no longer applies, but a more general theorem does:
Theorem — Banach–Nečas
Let $\mathcal U, \mathcal V$ be Hilbert spaces and $b_\mu$ a continuous bilinear form on $\mathcal U \times \mathcal V$. Suppose
- Inf-sup condition: there is $\beta_\mu > 0$ with $\displaystyle\sup_{v \in \mathcal V} \frac{b_\mu(u,v)}{|v|} \ge \beta_\mu |u|$ for all $u \in \mathcal U$, and
- No useless test functions: for every $v \ne 0$ in $\mathcal V$, there exists some $w \in \mathcal U$ with $b_\mu(w,v) \ne 0$.
Then for every $f_\mu \in \mathcal V’$, there exists a unique, stable $u_\mu \in \mathcal U$ with $b_\mu(u_\mu, v) = f_\mu(v)$ for all $v \in \mathcal V$.
Condition (1) gives stability and uniqueness (injectivity): if $u \ne 0$, some test function detects it, so the bilinear form can’t collapse two different solutions onto the same data. Condition (2) gives existence (surjectivity): no test function is “wasted,” so a solution can be found for any right-hand side.
Test yourself
Question: I have an ultraweak formulation for the wave equation. Can I use Lax–Milgram to prove it’s well-posed?
Answer: No. Ultraweak formulations typically use different trial and test spaces ($\mathcal U \ne \mathcal V$, a Petrov–Galerkin setup), and Lax–Milgram requires $\mathcal U = \mathcal V$. The wave equation is also generally not coercive in the relevant space-time norm. The right tool here is Banach–Nečas: you’d need to verify the inf-sup condition and the “no useless test function” condition instead.
Why any of this matters for model order reduction
💡 Why we care so much about αμ and βμ
This is where well-posedness theory stops being abstract functional analysis and starts driving the actual numerics of reduced basis methods.
Once you replace the true solution $u_\mu$ with a reduced approximation $u_\mu^\delta$, you need to know how far off you are — without knowing $u_\mu$ itself (that’s the entire point of a cheap online phase). This is exactly what motivates a-posteriori error estimation: bound the true error using only quantities you can actually compute from the approximation.
A typical residual-based bound looks like
\[\|u_\mu - u_\mu^\delta\|_{\mathcal U} \;\le\; \frac{1}{\beta_\mu}\, \|r_\mu^\delta\|_{\mathcal V'},\]where $r_\mu^\delta = f_\mu - B_\mu u_\mu^\delta$ is the residual — cheaply computable from the reduced solution alone.
Look at where $\beta_\mu$ sits: in the denominator.
- If we don’t know that $\beta_\mu > 0$, we can’t guarantee the error is even bounded.
- If $\beta_\mu$ gets close to $0$, the bound blows up, and the reduced model becomes untrustworthy — even if the actual error is small.
So a large part of model order reduction (the Successive Constraint Method, introduced later, is built entirely around this) is dedicated to computing cheap, tight, guaranteed lower bounds for $\alpha_\mu$ or $\beta_\mu$, in the same offline/online spirit as the affine decomposition from the last post: expensive work done once offline, cheap evaluation online for every new $\mu$.
This is also exactly why the whole coercivity / inf-sup / Lax–Milgram / Banach–Nečas discussion above isn’t just bureaucratic rigor — the constants that fall out of these theorems are the same constants that make or break the reliability of a reduced-order model.
Supremizers: making the inf-sup constant computable
Banach–Nečas tells us we need $\beta_\mu > 0$. It doesn’t tell us how to find it. The next piece of the theory turns “find the inf-sup constant” from an abstract search into something you can actually compute — and, later in the course, into something a reduced model can be built to preserve.
Definition — Supremizer Let $\mu \in \mathcal P$ and $0 \ne u \in \mathcal U$. The element $s_u(\mu) \in \mathcal V$ is called the supremizer of $u$ if
\(\|s_u(\mu)\|_{\mathcal V} = \arg\sup_{v \in \mathcal V} \frac{b_\mu(u,v)}{\|v\|_{\mathcal V}}.\)
In words: $s_u(\mu)$ is the “perfect partner” test function for $u$ — the specific $v$ that realizes the supremum in the inf-sup ratio. Plug this particular $v$ in, and the fraction $b_\mu(u,v)/|v|_{\mathcal V}$ hits its maximum possible value for that $u$.
Lemma 1.2.9 Let $\mu \in \mathcal P$ and $0 \ne u \in \mathcal U$. The supremizer $s_u(\mu)$ is the unique solution of
\((s_u(\mu), v)_{\mathcal V} = b_\mu(u,v) \qquad \text{for all } v \in \mathcal V.\)
Proof — the supremizer solves a linear variational problem
Existence and uniqueness. Consider the auxiliary problem: find \(w \in \mathcal V\) such that \((w,v)_{\mathcal V} = b_\mu(u,v)\) for all \(v \in \mathcal V\). The bilinear form here is the \(\mathcal V\)-inner product itself, which is automatically continuous (Cauchy–Schwarz) and automatically coercive with constant \(1\), since \((v,v)_{\mathcal V} = \|v\|_{\mathcal V}^2\) exactly. The right-hand side \(v \mapsto b_\mu(u,v)\) is a bounded linear functional because \(b_\mu\) is continuous. Lax–Milgram therefore gives a unique solution — call it \(s_u(\mu)\).
Matching the arg-sup definition. It remains to check this solution’s norm equals the supremum in Definition 1.2.8. By Cauchy–Schwarz, for any fixed $w \in \mathcal V$,
\[\sup_{v \in \mathcal V} \frac{(w,v)_{\mathcal V}}{\|v\|_{\mathcal V}} = \|w\|_{\mathcal V},\]with equality at $v = w$. Substituting $w = s_u(\mu)$ and using the defining relation:
\[\sup_{v \in \mathcal V} \frac{b_\mu(u,v)}{\|v\|_{\mathcal V}} = \sup_{v \in \mathcal V} \frac{(s_u(\mu),v)_{\mathcal V}}{\|v\|_{\mathcal V}} = \|s_u(\mu)\|_{\mathcal V},\]which proves the claim.
This is a genuine computational shortcut: finding a supremum over an infinite-dimensional space sounds hopeless, but the lemma says you never have to search for it. You just solve one more linear, Lax–Milgram-friendly problem — a Riesz representation problem posed entirely in $\mathcal V$ — and the supremizer falls out directly.
From supremizers to the inf-sup constant
Lemma 1.2.10 \(\beta_\mu = \inf_{u \in \mathcal U} \frac{\|s_u(\mu)\|_{\mathcal V}}{\|u\|_{\mathcal U}}.\)
Proof — the inf-sup constant in terms of supremizers
Start from the definition of $\beta_\mu$ and substitute the identity from Lemma 1.2.9:
\[\beta_\mu = \inf_{u \in \mathcal U} \sup_{v \in \mathcal V} \frac{b_\mu(u,v)}{\|u\|_{\mathcal U}\|v\|_{\mathcal V}} = \inf_{u \in \mathcal U} \sup_{v \in \mathcal V} \frac{(s_u(\mu),v)_{\mathcal V}}{\|u\|_{\mathcal U}\|v\|_{\mathcal V}} = \inf_{u \in \mathcal U} \frac{\|s_u(\mu)\|_{\mathcal V}}{\|u\|_{\mathcal U}},\]which proves the assertion.
Once you can map any $u$ to its supremizer $s_u(\mu)$ (call this map the supremizing operator $S_\mu: \mathcal U \to \mathcal V$, $S_\mu u := s_u(\mu)$), $\beta_\mu$ is just the worst-case ratio of that map — how much $S_\mu$ can shrink a vector, in the worst case. That’s exactly the language of singular values.
The geometric picture: βμ and Cμ as singular values
Think of $S_\mu$ as a linear map that takes the unit ball of $\mathcal U$ and stretches it into some ellipsoid in $\mathcal V$. The shortest semi-axis of that ellipsoid is $\beta_\mu$; the longest is the continuity constant $C_\mu$. Drag the slider below toward “ill-conditioned” and watch $\beta_\mu$ collapse toward zero — and remember from the last section that the a-posteriori error bound scales like $1/\beta_\mu$, so this is precisely the geometric picture of a reduced model becoming unreliable.
Drag right to make the problem less stable: the shortest axis of the ellipse (βμ) shrinks toward zero while the longest axis (Cμ) grows.
Formally: consider the generalized eigenvalue problem of finding orthonormal (generalized) eigenvectors \(\xi_i(\mu) \in \mathcal U\), \((\xi_i(\mu),\xi_j(\mu))_{\mathcal U} = \delta_{ij}\), and eigenvalues \(\lambda_i(\mu) \in \mathbb R\) such that
\[(S_\mu \xi_i(\mu), S_\mu w)_{\mathcal V} = \lambda_i(\mu)\,(\xi_i(\mu), w)_{\mathcal U} \qquad \text{for all } w \in \mathcal U.\]Proof — βμ and Cμ as extremal square roots of the eigenvalues
Since \(\|s_u(\mu)\|_{\mathcal V}^2 = \|S_\mu u\|_{\mathcal V}^2 = (S_\mu u, S_\mu u)_{\mathcal V}\), and the eigenvectors \(\xi_i(\mu)\) form an orthonormal basis of \(\mathcal U\), expand any \(u\) in this basis. Plugging \(u = \xi_j(\mu)\) into the eigenvalue relation gives
\[(S_\mu\xi_i(\mu), S_\mu\xi_j(\mu))_{\mathcal V} = \lambda_i(\mu)\,(\xi_i(\mu),\xi_j(\mu))_{\mathcal U} = \delta_{ij}\,\lambda_i(\mu).\]So $S_\mu$ acts diagonally in this basis, and
\[\beta_\mu^2 = \inf_{u \in \mathcal U} \frac{\|s_u(\mu)\|_{\mathcal V}^2}{\|u\|_{\mathcal U}^2} = \inf_{u \in \mathcal U} \frac{(S_\mu u, S_\mu u)_{\mathcal V}}{(u,u)_{\mathcal U}} = \inf_{i \in \mathbb N} (S_\mu\xi_i(\mu), S_\mu\xi_i(\mu))_{\mathcal V} = \inf_{i \in \mathbb N} \lambda_i(\mu),\]so that $\beta_\mu = \inf_{i \in \mathbb N} \sqrt{\lambda_i(\mu)}$. The same argument with $\inf$ replaced by $\sup$ gives the continuity constant, $C_\mu = \sup_{i \in \mathbb N} \sqrt{\lambda_i(\mu)}$.
So $\beta_\mu$ is exactly the square root of the smallest eigenvalue, and $C_\mu$ the square root of the largest — the same relationship a singular value decomposition has to the shortest and longest axes of the image of a unit ball, which is exactly the picture above.
Putting it in matrix form
💡 The algebraic version — what you'd actually code
Once $\mathcal U$ and $\mathcal V$ are discretized into finite bases $\Phi = {\varphi_i}$ and $\Psi = {\psi_i}$, every abstract object above becomes a concrete matrix or vector:
- \(\boldsymbol U = [(\varphi_i,\varphi_j)_{\mathcal U}]_{i,j}\) and \(\boldsymbol V = [(\psi_i,\psi_j)_{\mathcal V}]_{i,j}\) are the mass (Gramian) matrices representing the two inner products.
- \(\boldsymbol B_\mu^\top = [b_\mu(\varphi_i,\psi_j)]_{i,j}\) is the stiffness matrix representing the bilinear form. The supremizer equation from Lemma 1.2.9 becomes the linear system
and substituting this into the Rayleigh quotient for $\beta_\mu^2$ gives the algebraic form of the generalized eigenvalue problem:
\[\boldsymbol B_\mu^\top \boldsymbol V^{-1}\boldsymbol B_\mu\, \mathbf u = \lambda\, \boldsymbol U \mathbf u.\]In a discrete setting — for the full “truth” system as well as the reduced one — this can be solved with an SVD or any suitable generalized eigenvalue solver, to get $\beta_\mu = \sqrt{\lambda_{\min}}$ and $C_\mu = \sqrt{\lambda_{\max}}$ directly.
Why this matters later: once you build a reduced basis model for a non-coercive problem, projecting onto a small subspace can silently destroy the inf-sup stability you started with. The fix used later in the course is to explicitly compute supremizers for the reduced trial basis and enrich the reduced test space with them — keeping $\beta_\mu^\delta$ safely away from zero, and the error bound from the last section safely away from useless.
Discretization: Moving to the finite-dimensional world
Everything discussed so far applies to the exact, infinite-dimensional continuous problem. In practice, we solve PDEs on computers using finite-dimensional approximations, like the Finite Element Method.
Let’s say we introduce a discretization parameter $\delta$ (like a mesh size). We replace our infinite-dimensional spaces $\mathcal{U}$ and $\mathcal{V}$ with finite-dimensional subspaces $\mathcal{U}^\delta \subset \mathcal{U}$ and $\mathcal{V}^\delta \subset \mathcal{V}$. Our goal is to find a “truth” approximation $u_\mu^\delta \in \mathcal{U}^\delta$ such that: \(b_\mu(u_\mu^\delta, v^\delta) = f_\mu(v^\delta) \qquad \text{for all } v^\delta \in \mathcal{V}^\delta.\)
But there is a catch: Just because the continuous problem is stable, it does not automatically mean the discrete problem is stable. If we pick our spaces $\mathcal{U}^\delta$ and $\mathcal{V}^\delta$ poorly, our numerical approximation might blow up.
The LBB Condition: Discrete Inf-Sup
To ensure our discrete problem is well-posed, we must explicitly demand that our finite-dimensional spaces also satisfy an inf-sup condition. This is historically known as the LBB condition.
Definition 1.3.3 — The LBB Condition
The discrete spaces $\mathcal{U}^\delta \subset \mathcal{U}$ and $\mathcal{V}^\delta \subset \mathcal{V}$ satisfy the Ladyzhenskaya-Babuška-Brezzi (LBB) condition if there exists a discrete inf-sup constant $\beta_\mu^\circ > 0$ such that:
\[\inf_{u^\delta \in \mathcal{U}^\delta}\; \sup_{v^\delta \in \mathcal{V}^\delta} \; \frac{b(u^\delta, v^\delta; \mu)}{\|u^\delta\|_{\mathcal{U}}\|v^\delta\|_{\mathcal{V}}} \;\ge\; \beta_\mu^\circ\]If there exists a lower bound $\underline{\beta}^\circ > 0$ such that $\beta_\mu^\circ \ge \underline{\beta}^\circ$ for all parameters $\mu \in \mathcal{P}$ and all mesh sizes $\delta$, we say the uniform LBB condition holds.
Notice the little circle ($\circ$) notation! While $\beta_\mu$ is the continuous inf-sup constant, $\beta_\mu^\circ$ is the discrete inf-sup constant. We desperately need this to be strictly greater than zero to trust our numerical solver.
💡 A free pass for coercive problems
If you are dealing with a coercive problem (like the thermal block) and you use a standard Galerkin method where trial and test spaces are identical ($\mathcal{U}^\delta = \mathcal{V}^\delta$), you are in luck.
As stated in Remark 1.3.4, for Galerkin methods, the stability of discrete systems is automatically inherited from the coercivity of the continuous problem. You get $\beta_\mu^\circ = \alpha_\mu$ for free, meaning your discrete problem is unconditionally stable.
If your problem is only inf-sup stable (like the wave equation), this inheritance does not happen automatically. You have to be extremely careful how you choose your discrete spaces to satisfy LBB!
Xu-Zikatanov: The Generalized Céa’s Lemma
Once we establish that our discrete spaces are LBB-stable, we naturally want to know: “How accurate is our computed approximation $u_\mu^\delta$ compared to the true, exact solution $u_\mu$?”
The answer is given by a beautiful theorem that separates the concept of approximation power from the concept of numerical stability.
Theorem 1.3.5 — Xu-Zikatanov (2003)
Let $b_\mu$ be a continuous bilinear form with continuity constant $C_\mu$. Suppose the LBB condition holds for our discrete spaces with constant $\beta_\mu^\circ > 0$. Then, the exact solution $u_\mu$ and the discrete approximation $u_\mu^\delta$ satisfy the estimate:
\(\|u_\mu - u_\mu^\delta\|_{\mathcal U} \;\le\; \frac{C_\mu}{\beta_\mu^\circ} \inf_{w^\delta \in \mathcal{U}^\delta} \|u_\mu - w^\delta\|_{\mathcal U}\)
This formula tells a powerful story. The term on the far right ($\inf_{w^\delta}$) represents the absolute best possible approximation you could achieve if an oracle handed you the perfect function from your finite-dimensional space.
The fraction $\frac{C_\mu}{\beta_\mu^\circ}$ is exactly the condition number $\kappa$ we discussed in the geometry section above. Therefore, the theorem states that your computed error is bounded by the best possible error, scaled by the condition number of your discrete setup. If your discrete inf-sup constant $\beta_\mu^\circ$ is healthy, your computed solution is basically as good as the oracle’s!
Céa’s Lemma: The Coercive Case
If we restrict ourselves to the easier, well-behaved world of coercive problems, the Xu-Zikatanov theorem simplifies into one of the most famous results in numerical analysis.
Corollary 1.3.6 — Céa’s Lemma
Let the conditions of the Lax-Milgram theorem hold (i.e., the problem is coercive with constant $\alpha_\mu$). If $\mathcal{U}^\delta \subset \mathcal{U}$ is finite-dimensional, then:
\(\|u_\mu - u_\mu^\delta\|_{\mathcal U} \;\le\; \frac{C_\mu}{\alpha_\mu} \inf_{w^\delta \in \mathcal{U}^\delta} \|u_\mu - w^\delta\|_{\mathcal U}\)
Because coercivity automatically gives us LBB stability with $\beta_\mu^\circ = \alpha_\mu$, we simply swap the constants.
Why this matters for Model Reduction
When we transition to building Reduced Basis Models in Chapter 1.4, we are going to repeat this exact same process. Instead of projecting from the continuous world to the “truth” finite element world, we will project from the “truth” space to a tiny, $N$-dimensional reduced space $\mathcal{U}^N$.
To ensure our incredibly fast reduced models don’t spit out garbage, we will have to guarantee that the LBB condition still holds in our reduced spaces, ensuring a healthy $\beta_\mu^N > 0$.
This page builds on the lecture Model Reduction by Prof. Urban, Summer Semester 2026.