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For a continuously differentiable convex function on ${\mathbb{R}}^n$ with coordinate Lipschitz continuity constants $L_i$, its block coordinate gradient is Lipschitz continuous along each corresponding coordinate subspace, with constant at most $S_{\mathcal{B}}={\sum }_{i\in \mathcal{B}}{L}_i$.

To prove this without assuming a Hessian matrix, put ${\delta }_{\mathcal{B}}={\nabla }_{\mathcal{B}}f(x)-{\nabla }_{\mathcal{B}}f(y)$. Sum the coordinate derivative bound by Bregman divergence over the block, then repeat with $x,y$ interchanged and average:If $x-y$ is supported in the block, the dot product uses only ${\delta }_{\mathcal{B}}$. The Cauchy-Schwarz inequality and cancellation of its Euclidean norm give the result, including the zero case separately. Taking the full block supplies a global Lipschitz gradient constant ${\sum }_i{L}_i$.

Target: convex-optimization.bigb, under Coordinate descent.

Suppose $f$ is a differentiable convex function with coordinate Lipschitz continuity constant $L_i>0$. Put $\varphi (u)=D_f(u,y)=f(u)-f(y)-\nabla f(y{)}^T(u-y)$, the Bregman divergence from the supporting affine function at $y$. The subgradient inequality gives $\varphi \ge 0$, and its coordinate Lipschitz constant is still $L_i$. The coordinate descent lemma with step size $1/L_i$ yieldsSince ${\partial }_i\varphi (x)={\partial }_{i}f(x)-{\partial }_{i}f(y)$, rearrangement proves the bound. Continuous differentiability suffices; a Hessian matrix is unnecessary.

Target: gradient-descent.bigb, under Lipschitz gradient.

The subgradient inequality for the differentiable convex function $f$ givesThus $y$ is a global minimizer of $\varphi$ with value zero. Subtracting an affine function preserves convexity.

We must also justify using part (a) for $\varphi$, since its partial derivative can differ from that of $f$. First show that condition (6) for a singleton block implies the full coordinate Lipschitz continuity estimateIf ${\partial }_{i}f(z)\ne 0$, choose $\alpha =-t/{\partial }_{i}f(z)$ in (6). If it is zero but ${\partial }_{i}f(z+t{e}_i)\ne 0$, apply the same argument starting at $z+t{e}_i$ and reverse the displacement. If both partial derivatives are zero, the estimate is immediate. The availability of all real $\alpha$ in (6) justifies these choices.

Now ${\partial }_i\varphi (z)={\partial }_{i}f(z)-{\partial }_{i}f(y)$, so differences of this partial derivative along coordinate lines are unchanged. Hence $\varphi$ has the same coordinate Lipschitz continuity constant $L_i$ and satisfies the singleton version of (6). Apply part (a) to $\varphi$ with step size $1/L_i$:Combining with its already established nonnegativity provesThe coordinate estimate survives subtraction of the supporting affine function. This is the step that allows the arbitrary comparison point $y$ to be exploited in part (c).

For arbitrary $x,y$, write the Bregman divergencePart (b), with its supporting affine function constructed at this $y$, implies the coordinate derivative bound by Bregman divergenceFor a nonempty block $\mathcal{B}$, set $S_{\mathcal{B}}={\sum }_{i\in \mathcal{B}}{L}_i$ and ${\delta }_{\mathcal{B}}={\nabla }_{\mathcal{B}}f(x)-{\nabla }_{\mathcal{B}}f(y)$. Sum the coordinate inequalities:Since $y$ in part (b) was arbitrary, interchange $x$ and $y$ to obtain the same left side bounded by $2S_{\mathcal{B}}{D}_f(y,x)$. Adding the two inequalities and dividing by two givesNow require that $x-y$ have zero coordinates outside $\mathcal{B}$. The last dot product is then ${\delta }_{\mathcal{B}}^T(x-y)$. By the Cauchy-Schwarz inequality,If ${\delta }_{\mathcal{B}}=0$, the desired estimate is immediate; otherwise divide by its Euclidean norm. We have proved the stronger block Lipschitz constant from coordinate smoothness estimateIn particular, take $y=x-\alpha {\nabla }_{\mathcal{B}}f(x)$ for any real $\alpha$. This proves (6), with a valid choiceThe least admissible Lipschitz constant is therefore no larger than this choice. For the empty block, the block coordinate gradient vanishes identically and its least nonnegative Lipschitz constant is zero. The proof uses only $C^1$ differentiability, convexity and coordinate Lipschitz continuity, with no assumed Hessian matrix.