Question 1

Where does training end up?

Given two training procedures, datasets, objectives or initializations, what solutions do they approach, and how far apart are those solutions?

Symptoms

Is this your question?

You are probably here if your claim sounds like one of these:

  • Both methods recover the same solution.
  • Removing one training point changes the optimum by \(\mathcal{O}(1/n)\).
  • Training on all the data ends up close to training on part of it.
  • Gradient descent selects the max-margin or minimum-norm solution.
  • The solution is stable under a small change to the objective.
Sub-cases

Which version are you facing?

Case Usually start with
Exactly the same optimum? First-order or KKT conditions, with uniqueness from strict or strong convexity. If the minimizer is not unique, characterize the whole set.
A nearby optimum after a small perturbation? Strong-convexity sensitivity, the implicit function theorem, influence functions.
Different algorithms pick different points of the same minimizer set? Implicit-bias analysis: the dynamics, their invariants, and the KKT conditions of the problem you conjecture they solve in the limit.
The same fixed point under different update rules? Fixed-point equations and contraction arguments.
The recipe

Subtract the two optimality equations

Start from the equation that defines each endpoint, not from the training update. For an isolated optimum that is usually \(\nabla f(\theta^\star) = 0\); for a constrained problem it is the KKT system; for implicit bias it may be the KKT system of a different problem altogether, such as a max-margin or minimum-norm program.

Subtract the equations for the two setups. You almost always want something of the form

\[A\,(\theta_1 - \theta_2) = r,\]

or its nonlinear analogue. Lower-bound the left side with curvature or strong monotonicity, and upper-bound the residual \(r\) by the size of the perturbation. Only then translate parameter distance into distance in predictions, loss or representations.

When there is no finite optimum, change what you track. Implicit-bias questions need a different route:

  1. Show the parameters do not converge. For example, on separable data the logistic loss tends to zero only as \(\lVert \theta_t \rVert \to \infty\).
  2. Pick the object that does converge: the direction \(\theta_t / \lVert \theta_t \rVert\), a normalized predictor, the margin or the rank.
  3. Guess the limiting problem (hard-margin SVM, minimum norm, minimum nuclear norm) and write its KKT conditions.
  4. Show the normalized dynamics satisfy those conditions in the limit. With an exponential-tailed loss, the gradient becomes dominated by the examples with the smallest margin, which play the role of support vectors.
A complete proof

How far does a small extra term move the minimizer?

The claim

Let \(F_\varepsilon(\theta) = L_H(\theta) + \varepsilon\, L_T(\theta)\), with \(\theta_H = \arg\min L_H\) and \(\theta^\star_\varepsilon = \arg\min F_\varepsilon\). Then

\[\lVert \theta^\star_\varepsilon - \theta_H \rVert \;\le\; \frac{\varepsilon\, \lVert \nabla L_T(\theta_H) \rVert}{\mu}.\]
Assumptions
  • \(L_H\) is differentiable and \(\mu\)-strongly convex.
  • \(L_T\) is differentiable and convex.
  • \(\varepsilon \ge 0\), so \(F_\varepsilon\) is strongly convex and both minimizers exist and are unique.
First move, and why

Write both optimality conditions. We want a statement about where two minimizers sit, and the only thing we know exactly about a minimizer is that its gradient vanishes. Strong convexity is the tool that turns a gap in gradients into a gap in position.

Show the proof, with commentary

Step 1. Let \(d = \theta^\star_\varepsilon - \theta_H\). Optimality gives \(\nabla L_H(\theta^\star_\varepsilon) = -\varepsilon \nabla L_T(\theta^\star_\varepsilon)\) and \(\nabla L_H(\theta_H) = 0\).

Step 2. Strong convexity of \(L_H\) means its gradient is strongly monotone:

\[\mu \lVert d \rVert^2 \le \langle \nabla L_H(\theta^\star_\varepsilon) - \nabla L_H(\theta_H), d \rangle = -\varepsilon \langle \nabla L_T(\theta^\star_\varepsilon), d \rangle .\]

This is the subtraction from the recipe. The right side is evaluated at the unknown point \(\theta^\star_\varepsilon\), which is the awkward part.

Step 3. Move the gradient of \(L_T\) back to a point we know. Convexity of \(L_T\) makes its gradient monotone, \(\langle \nabla L_T(\theta^\star_\varepsilon) - \nabla L_T(\theta_H), d \rangle \ge 0\), so

\[\mu \lVert d \rVert^2 \le -\varepsilon \langle \nabla L_T(\theta_H), d \rangle \le \varepsilon \lVert \nabla L_T(\theta_H) \rVert\, \lVert d \rVert ,\]

using Cauchy–Schwarz for the second inequality.

This is the step that uses convexity of \(L_T\). Without it you can still finish, but you need a bound on \(\nabla L_T\) over the whole region where \(\theta^\star_\varepsilon\) could be.

Step 4. If \(d = 0\) there is nothing to prove; otherwise divide by \(\lVert d \rVert\). \(\blacksquare\)

A tempting approach that fails

Taylor-expand the optimality condition around \(\theta_H\) and solve: \(\theta^\star_\varepsilon \approx \theta_H - \varepsilon\, \nabla^2 L_H(\theta_H)^{-1} \nabla L_T(\theta_H)\). This is the implicit function theorem, which also needs \(L_H\) twice differentiable with an invertible Hessian, and it even gets the leading constant right. But it is a statement about the limit \(\varepsilon \to 0\): the remainder is \(o(\varepsilon)\) with no explicit size, so it proves nothing for the \(\varepsilon\) you actually have. Turning it into a bound needs extra assumptions, such as a Lipschitz Hessian on a ball that you have already shown contains \(\theta^\star_\varepsilon\). The monotonicity argument above needs neither.

A second route, and when to prefer it

Compare function values instead of gradients. If \(f\) and \(g\) are both \(\mu\)-strongly convex, with minimizers \(x_f\) and \(x_g\), and \(\lvert f - g \rvert \le \delta\) at those two points, then

\[\tfrac{\mu}{2}\lVert x_f - x_g \rVert^2 \le f(x_g) - f(x_f), \qquad \tfrac{\mu}{2}\lVert x_f - x_g \rVert^2 \le g(x_f) - g(x_g).\]

Adding the two and regrouping gives \(\mu \lVert x_f - x_g \rVert^2 \le \big(f(x_g) - g(x_g)\big) + \big(g(x_f) - f(x_f)\big) \le 2\delta\), so \(\lVert x_f - x_g \rVert \le \sqrt{2\delta/\mu}\).

With \(\delta\) proportional to \(\varepsilon\), this gives \(\mathcal{O}(\sqrt{\varepsilon/\mu})\), a weaker rate than the \(\mathcal{O}(\varepsilon/\mu)\) of the gradient route. In exchange it asks for less: no gradient of \(L_T\), only a bound on how far the two losses can differ. And its key step, "a small loss gap forces a small distance", is exactly what a Hölder error bound provides when strong convexity fails. That is why it is the route that carries over to deep networks, and the one the CLTR paper takes.

If you relax an assumption
  • \(L_T\) not convex: for a minimizer \(\theta^\star_\varepsilon\) (unique once \(\varepsilon\) times the smoothness of \(L_T\) is below \(\mu\)), Step 2 still gives \(\lVert d \rVert \le \varepsilon \lVert \nabla L_T(\theta^\star_\varepsilon) \rVert / \mu\), so you need a gradient bound \(G\) on the region that can contain \(\theta^\star_\varepsilon\), and the result becomes \(\varepsilon G/\mu\).
  • \(L_H\) only strongly convex near \(\theta_H\): first prove \(\theta^\star_\varepsilon\) stays in that neighborhood, then run the same argument there.
  • No strong convexity at all, as in deep networks: assume instead a local error bound \(\mathrm{dist}(\theta, S_H) \le C\sqrt{L_H(\theta) - \min L_H}\) near the set \(S_H\) of minimizers of \(L_H\), and bound the distance to that set. A KL inequality with exponent \(\tfrac12\) at the minimizers implies the bound, but for a given architecture neither comes for free. Combined with the second route, this is the structure of Theorem 3.5 of the CLTR paper.
0.13
How much the second term counts.
0.50
Flatter valleys let the solution drift further.
\(\theta_H\), minimizer of \(L_H\) minimizer of \(L_T\) \(\theta^\star_\varepsilon\) bound from the lemma
The lemma, live. Blue contours are \(L_H\), here a quadratic with curvature \(\mu\) along its flattest direction. The second term pulls toward the orange point. The purple curve is every \(\theta^\star_\varepsilon\) as \(\varepsilon\) runs from 0 to 10, and the dashed circle has the radius the lemma allows. Lower \(\mu\) and the circle grows, because a flat valley offers little resistance.
Try it yourself

Ridge regression with an extra pull

The exercise

Let \(L_H(\theta) = \tfrac12\lVert A\theta - b \rVert^2 + \tfrac{\lambda}{2}\lVert \theta \rVert^2\) with \(\lambda > 0\), and add \(\varepsilon L_T\) with \(L_T(\theta) = \tfrac12\lVert \theta - c \rVert^2\).

  1. Use the lemma above to bound \(\lVert \theta^\star_\varepsilon - \theta_H \rVert\).
  2. Solve for \(\theta^\star_\varepsilon\) exactly and compute the true distance.
  3. When is the lemma's bound nearly tight, and by how much can it overshoot?
Hint

\(L_H\) is \(\mu\)-strongly convex with \(\mu\) equal to the smallest eigenvalue of \(M = A^\top A + \lambda I\). For the exact solution, set the gradient of \(L_H + \varepsilon L_T\) to zero and subtract \(M\theta_H = A^\top b\).

Show a solution

1. \(\mu = \lambda_{\min}(M) \ge \lambda\) and \(\nabla L_T(\theta_H) = \theta_H - c\), so the lemma gives \(\lVert \theta^\star_\varepsilon - \theta_H \rVert \le \varepsilon \lVert \theta_H - c \rVert / \mu\).

2. Optimality gives \((M + \varepsilon I)\,\theta^\star_\varepsilon = A^\top b + \varepsilon c\). Subtracting \((M + \varepsilon I)\theta_H = A^\top b + \varepsilon \theta_H\),

\[\theta^\star_\varepsilon - \theta_H = \varepsilon\,(M + \varepsilon I)^{-1}(c - \theta_H), \qquad \lVert \theta^\star_\varepsilon - \theta_H \rVert \le \frac{\varepsilon\,\lVert c - \theta_H \rVert}{\mu + \varepsilon}.\]

3. The exact distance equals \(\varepsilon\lVert c - \theta_H \rVert/(\mu + \varepsilon)\) when \(c - \theta_H\) points along the flattest direction of \(M\). Then the lemma overshoots only by the factor \((\mu + \varepsilon)/\mu\), which tends to 1 as \(\varepsilon \to 0\). If \(c - \theta_H\) points along a steep direction, with eigenvalue \(\Lambda\), the truth is smaller by roughly \(\mu/\Lambda\): the lemma pays for the flattest direction whether or not the perturbation uses it.

The toolkit

What each tool gives you, and what it costs

First-order and KKT conditions

\[\nabla f(x^\star) + \textstyle\sum_i \lambda_i \nabla g_i(x^\star) + A^\top \nu = 0,\quad \lambda_i \ge 0,\quad \lambda_i g_i(x^\star) = 0\]

For a convex problem \(\min f(x)\) subject to \(g_i(x) \le 0,\ Ax = b\) under Slater's condition, plus primal feasibility. Unconstrained: an interior local minimum of a differentiable \(f\) has \(\nabla f(x^\star) = 0\).

Gives
Turns "which solution?" into algebra, and often reveals an implicit regularizer.
Costs
Differentiability and a constraint qualification; sufficiency needs convexity or another global argument.
Wrong tool when
The problem is non-convex: a stationary point need not be a minimum, and KKT says nothing about which one the dynamics pick.

Strong-convexity sensitivity

\[\sup_x \lVert \nabla f(x) - \nabla \tilde f(x) \rVert \le \varepsilon \;\Longrightarrow\; \lVert \tilde x^\star - x^\star \rVert \le \frac{\varepsilon}{\mu}\]

For differentiable, \(\mu\)-strongly convex \(f\), with \(x^\star = \arg\min f\) and \(\tilde x^\star = \arg\min \tilde f\).

Gives
A direct bound on how far the parameters move.
Costs
A unique minimizer and curvature \(\mu > 0\), at least on the region that matters.
Wrong tool when
Minima are flat or overparameterized, or symmetries make Euclidean parameter distance meaningless.

Uniform objective perturbation

\[\sup_x \lvert f(x) - \tilde f(x) \rvert \le \delta \;\Longrightarrow\; \lVert \tilde x^\star - x^\star \rVert \le 2\sqrt{\delta/\mu}\]

For \(\mu\)-strongly convex \(f\).

Gives
Stability when you can compare only function values, not gradients.
Costs
Strong convexity and a uniform approximation on the relevant domain.
Wrong tool when
You do know the gradient perturbation; the previous tool gives the sharper \(\mathcal{O}(\varepsilon/\mu)\).

Implicit function theorem and influence functions

\[\left.\frac{d\theta^\star_\epsilon}{d\epsilon}\right|_{0} = -H^{-1}\, \partial_\epsilon F(\theta^\star, \epsilon)\big|_{0}, \qquad \text{upweighting } z:\ \frac{d\theta^\star}{d\epsilon} = -H^{-1}\nabla_\theta \ell(z, \theta^\star)\]

With \(F(\theta, \epsilon) = \nabla_\theta R(\theta, \epsilon)\), \(F(\theta^\star, 0) = 0\) and \(H = \partial_\theta F(\theta^\star, 0)\) invertible.

Gives
The first-order effect of reweighting or deleting data, or of changing a hyperparameter.
Costs
Local differentiability and a nonsingular Hessian; the answer holds only for small \(\epsilon\).
Wrong tool when
The Hessian is singular, the perturbation is large, or the optimum is non-smooth.

Contraction mappings

\[d(x_t, x^\star) \le q^t\, d(x_0, x^\star), \qquad 0 \le q < 1\]

If \(T\) is a \(q\)-contraction on a complete metric space, it has a unique fixed point \(x^\star\).

Gives
Uniqueness, convergence to the equilibrium and its sensitivity, all at once.
Costs
A genuine contraction on an invariant complete set.
Wrong tool when
The dynamics are only non-expansive, or have large neutral directions, as in most neural-network training.

Kurdyka–Łojasiewicz (KL) inequality

\[\varphi'\big(f(x) - f(x^\star)\big)\, \operatorname{dist}\big(0, \partial f(x)\big) \ge 1 \quad \text{near } x^\star\]

With \(\varphi(0) = 0\) and \(\varphi' > 0\); exponent \(1/2\) corresponds to \(\varphi(s) = c\sqrt{s}\).

Gives
Convergence of descent methods and, with exponent \(1/2\) at the minimizers, a local error bound: distance to the set of minimizers in non-convex problems, where no unique optimum exists.
Costs
Losses built from analytic or semialgebraic pieces have the KL property with some exponent; the exponent \(1/2\) and the neighborhood are extra assumptions. The bound is to a set, not a single point.
Wrong tool when
It is asserted abstractly with no link to the actual architecture or loss.
Worked examples

How published papers answer it

  1. Soudry, Hoffer, Nacson, Gunasekar & Srebro, The Implicit Bias of Gradient Descent on Separable Data, JMLR 2018. arXiv

    On linearly separable data with the logistic loss, the weights grow without bound while their direction converges to the hard-margin SVM direction. The proof analyzes the gradient-descent asymptotics and matches the limiting direction with max-margin optimality conditions. It is the standard warning that "where training ends up" need not be a finite point.

  2. Gunasekar, Woodworth, Bhojanapalli, Neyshabur & Srebro, Implicit Regularization in Matrix Factorization, NeurIPS 2017. arXiv

    Studies gradient descent on an overparameterized matrix factorization from small initialization, and relates the solution it selects to the minimum nuclear-norm solution: proved in a special case, conjectured in general. A clean template for identifying which member of a minimizer set the dynamics choose.

  3. Koh & Liang, Understanding Black-box Predictions via Influence Functions, ICML 2017. arXiv

    Differentiates the optimum after infinitesimally upweighting one training point, giving the \(-H^{-1}\nabla\ell\) formula above, and makes it practical with Hessian–vector products. The paper is also explicit about the gap between the convex, smooth assumptions behind influence functions and the non-convex networks it applies them to.

Common traps

Where proofs of this kind go wrong

Resources

Where to go next

← Back to the guide

Found an error, or have a better example or a question this guide should cover? Email m.molahasani.m@gmail.com. Corrections and contributions are welcome. Last updated October 5, 2026.