Slide 20 — Proof of KL Convergence (Smooth Case)

Theorem (Smooth KL Convergence).

Let $f \colon \mathbb{R}^n \to \mathbb{R}$ be $C^1$ with $L$-Lipschitz gradient, $\inf f > -\infty$, and suppose the sublevel set $\{x : f(x) \leq f(x^0)\}$ is bounded. Suppose further that $f$ satisfies the KL property at every critical point. Run gradient descent with step size $\alpha \leq 1/L$: $$x^{k+1} = x^k - \alpha\,\nabla f(x^k).$$ Then the sequence $\{x^k\}$ converges to a critical point of $f$.

Remark. The theorem above only guarantees convergence to a critical point, which could be a local minimum, local maximum, or saddle point. To rule out saddle points, we appeal to a separate result:

Theorem (Lee, Simchowitz, Jordan & Recht, SIAM J. Optim. 2017, Theorem 4.1). Let $f$ be $C^2$ with Lipschitz continuous gradient. Then the set of initializations from which gradient descent converges to a strict saddle point has Lebesgue measure zero.

Combined with the theorem above (see slide o2), gradient descent converges to a local minimizer for almost every initialization.

Warm-up: complete proof for the smooth case (gradient descent).

We prove convergence in the simplest setting: $f$ is $C^1$ with $L$-Lipschitz gradient, and the algorithm is gradient descent $x^{k+1} = x^k - \alpha\nabla f(x^k)$ with $\alpha \leq 1/L$. This avoids subdifferentials entirely and makes the role of KL transparent.

Assumption. $f \colon \mathbb{R}^n \to \mathbb{R}$ is $C^1$, $\nabla f$ is $L$-Lipschitz, $\inf f > -\infty$, the sublevel set $\{x : f(x) \leq f(x^0)\}$ is bounded, and $f$ satisfies the KL property at every critical point.

Step 1: Sufficient decrease. The descent lemma (slide r1) gives, using $\|x^{k+1} - x^k\| = \alpha\|\nabla f(x^k)\|$:

Step 2: Preliminary consequences. Summing Step 1 over $k = 0, 1, \ldots$:

where $f^* = \lim_{k\to\infty} f(x^k)$ exists since $f(x^k)$ is nonincreasing and bounded below. In particular, $\|x^{k+1}-x^k\| \to 0$ and $\|\nabla f(x^k)\| \to 0$. Since all iterates lie in the bounded sublevel set $\{f \leq f(x^0)\}$, Bolzano-Weierstrass gives a subsequential cluster point: $x^{k_j} \to \tilde{x}$ for some subsequence. By continuity, $f(\tilde{x}) = f^*$ and $\nabla f(\tilde{x}) = 0$.

Step 3: The KL inequality at $\tilde{x}$. Since $\nabla f(\tilde{x}) = 0$, the KL property applies at $\tilde{x}$: there exist $r > 0$, $\eta > 0$, and a concave $C^1$ function $\varphi\colon[0,\eta)\to[0,\infty)$ with $\varphi(0)=0$, $\varphi'>0$, such that

Step 4: Conditional telescoping bound. Define $\Delta_k = \varphi(f(x^k)-f^*)$. We claim:

Proof. By concavity of $\varphi$ (tangent line lies above the graph):

Now substitute the two key inequalities:

• Sufficient decrease (Step 1): $f(x^k)-f(x^{k+1}) \geq \tfrac{1}{2\alpha}\|x^{k+1}-x^k\|^2$.
• KL inequality (Step 3): $\varphi'(f(x^k)-f^*) \geq \tfrac{1}{\|\nabla f(x^k)\|} = \tfrac{\alpha}{\|x^{k+1}-x^k\|}$.

Step 5: Iterates eventually stay in the KL neighborhood.

Claim. There exists an index $K_0$ such that $x^k \in B(\tilde{x},r)$ and $f^* < f(x^k) < f^*+\eta$ for all $k \geq K_0$.

Proof. Since $x^{k_j}\to\tilde{x}$ and $f(x^k)\to f^*$ and $\varphi$ is continuous with $\varphi(0)=0$, choose $k_j$ large enough that simultaneously:

• $\|x^{k_j}-\tilde{x}\| < r/2$, so $x^{k_j}\in B(\tilde{x},r)$;
• $f^* < f(x^{k_j}) < f^*+\eta$; and
• $2\,\varphi(f(x^{k_j})-f^*) < r/2$.

The Claim asserts two conditions simultaneously: spatial ($x^k\in B(\tilde{x},r)$) and $f$-value ($f^* < f(x^k) < f^*+\eta$). We dismiss the $f$-value exit first:

• Upper bound $f(x^k) < f^*+\eta$: $f(x^k)$ is nonincreasing (Step 1), so $f(x^k)\leq f(x^{k_j}) < f^*+\eta$ for all $k\geq k_j$. The upper bound is never violated.
• Lower bound $f(x^k) \geq f^*$: $f(x^k)$ is nonincreasing and has a subsequence $f(x^{k_j})\to f^*$; hence the full sequence converges to $f^*$ from above, so $f(x^k)\geq f^*$ for all $k$. The value $f(x^k) < f^*$ is impossible. If $f(x^K)=f^*$ for some $K\geq k_j$, then the descent condition gives $f^*+\tfrac{1}{2\alpha}\|x^{K+1}-x^K\|^2\leq f(x^K)=f^*$, forcing $x^{K+1}=x^K$. The sequence is stationary from $K$ on and trivially stays in $B(\tilde{x},r)$; the Claim holds immediately.

So the only way the Claim can fail is through the spatial condition. Suppose for contradiction that some iterate leaves $B(\tilde{x},r)$ after $k_j$. Let $K^*>k_j$ be the first such index. For $k_j \leq k < K^*$, $x^k\in B(\tilde{x},r)$ by definition of $K^*$, and $f^* < f(x^k) < f^*+\eta$ by the two points above. So Step 4 applies at each such $k$, and summing:

But then by the triangle inequality:

a contradiction. So $K^*$ does not exist, and the Claim holds with $K_0 = k_j$. $\square$

Step 6: Convergence and criticality. By the Claim, Step 4 applies for all $k \geq K_0$. Summing and letting $M\to\infty$:

The sequence $\{x^k\}$ has finite total path length, hence is Cauchy in $\mathbb{R}^n$, and converges to some $x^*$. Since $\nabla f$ is continuous and $\|\nabla f(x^k)\|\to 0$, we conclude $\nabla f(x^*)=0$. $\square$

General case: proof of Theorem 2.9 (Attouch–Bolte–Svaiter 2013).

The smooth warm-up above is a special case of Theorem 2.9, which covers any algorithm satisfying H1–H3 with a proper l.s.c. KL objective. The full proof extends the warm-up by an index-shift trick (H2 at step $k-1$ bounds the subgradient at $x^k$) and a coupled recurrence ($\star$) that replaces the clean telescoping estimate. A self-contained proof with all six steps is given at slides_KL_nonsmooth.html.