Subgradients and supporting hyperplanes
Replace a derivative with global affine lower bounds and handle corners without guessing an ordinary slope.
On this page
The question for today
What you’ll understand
- Compute subdifferentials of absolute values and finite maxima.
- Prove the zero-subgradient optimality condition and interpret normal cones.
- Identify where boundary subgradients can fail to exist.
At a corner, asking for the derivative demands a single slope that the graph cannot provide. A subgradient asks a different question: which slopes define affine functions lying below the entire objective and touching it at the chosen point? At a smooth convex point there is one such slope, while a corner may support an interval of them. Because the inequality is global, a zero subgradient certifies global optimality immediately. The same definition also explains constraint forces through normal cones and exposes why finite boundary values need not have supporting slopes.
A slope must support the whole function
For a proper convex f on and a finite-domain point x, a vector g is a subgradient when its affine prediction at x is no larger than f(z) for every z. The collection is the subdifferential. Outside the domain we define it to be empty. A differentiable convex function has subdifferential consisting only of its gradient, because directional derivatives force equality of every component. For |x|, the slope is −1 on the negative side, +1 on the positive side, and any value in [−1,1] at zero.
Subgradients are not arbitrary local trend estimates. A line that sits below a graph near x but crosses above it elsewhere fails the definition. For convex functions, local support around an interior point extends globally, but that conclusion uses convexity. Adding the two subgradient inequalities at x and y also shows monotonicity: the difference of supporting slopes has nonnegative inner product with x−y. In one dimension this generalizes the familiar statement that the derivative of a convex function is nondecreasing, now allowing vertical intervals of slopes at corners.
Optimality and constraint normals
The condition zero belongs to the subdifferential is equivalent to x being a global minimizer, directly by setting g=0 in the definition. For a convex indicator, its subdifferential is the normal cone: vectors whose inner product with every feasible displacement is nonpositive. At an interior point of a full-dimensional convex set, that cone contains only zero. At a boundary it can contain outward-pointing vectors. A constrained optimum balances the loss gradient against a normal; the gradient alone generally does not vanish at a boundary solution.
When h is convex and differentiable everywhere and C is nonempty closed convex, the sum rule gives the subdifferential of h plus the indicator as gradient h plus the normal cone. More generally, sums of proper convex functions need a domain qualification for equality of subdifferential sums; meeting relative interiors is a standard finite-dimensional sufficient condition. For a maximum of finitely many affine functions, the subdifferential at a point is the convex hull of the active slopes. Inactive pieces do not touch the graph and cannot contribute to its supporting slope set there.
Boundary support can disappear
In finite dimensions a proper convex function has at least one subgradient at every point in the relative interior of its effective domain. At relative-boundary points this can fail even for a closed convex function. Take f(x)=−√x for x≥0 and infinity otherwise. A finite subgradient g at zero would require −√z≥gz for every positive z, hence g≤−1/√z for arbitrarily small z. No finite g satisfies that requirement. The epigraph has no supporting hyperplane with the nonvertical orientation required by a finite slope there.
This example prevents an unsafe recipe: one cannot always pick a subgradient at any finite point and run an algorithm. The evaluation oracle itself needs a domain guarantee. It also clarifies the role of the relative interior in calculus rules and duality qualifications. Finally, subgradients are exact mathematical lower bounds; inexact numerical slopes require an explicit tolerance and a modified inequality. Treating a noisy estimate as a certified support can produce a false lower bound or an incorrect stopping decision, even when the underlying function really is convex.
Work it through
A minimum at a crossing of affine pieces
For f(x)=max{2x−1,−x+2}, compute its subdifferential and minimize it.
The pieces cross where 2x−1=−x+2, giving x=1 and value 1. To the left, −x+2 is larger and has slope −1; to the right, 2x−1 is larger and has slope 2.
At the crossing, the active-slope rule gives ∂f(1)=[−1,2]. Since zero lies in this interval, the crossing is a global minimizer.
Every point to the left has cost 2−x>1, and every point to the right has cost 2x−1>1. This also proves the minimum is unique, even though f is not strictly convex on either side.
The unique minimizer is x=1 with cost 1. The corner supplies many subgradients, including the zero vector that certifies optimality.
The living notebook
Explore the model
Interactive lab · function
One smooth supporting slope
Pause and Reset, then set a=0.5,x=1. Predict y=0.5 and tangent slope 1. Check that the tangent stays below the full quadratic: its gap at any z is (z−1)²/2. For a fresh comparison Reset and set a=−0.5,x=1; the tangent is now an upper support, so it is not a convex subgradient bound. This model has smooth quadratics only and does not render the interval of slopes at a corner.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe support inequality, rather than the existence of a drawn tangent, determines whether a slope is a convex subgradient.
Pause & reason
Check your understanding
Your turn
Try it yourself
Minimize F(x)=|x|+(x−2)²/2 on R by checking its subdifferential, including x=0.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
On each side of zero, add the absolute-value slope to x−2. At zero, add −2 to the interval [−1,1].
Compare with the solution
For x>0, the slope is 1+x−2=x−1, which vanishes at x=1. For x<0, it is −1+x−2=x−3, whose zero is not in that region. At zero the subdifferential is [−3,−1], which excludes zero. Thus x=1 is the minimizer, with cost 1+1/2=3/2. The quadratic term is strictly convex, so adding the convex absolute value preserves strict convexity and guarantees uniqueness.
Keep the idea
Take it with you
- Subgradients are global supporting affine slopes, including sets of slopes at corners.
- A zero subgradient is exactly a certificate of global minimum.
- Finite-dimensional relative-interior points have subgradients; boundary points may not.
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