Convex and Semicontinuous Functions on R^n: Examples and Counterexamples

Homogeneity and the triangle inequality in [L1] give $\mathrm{\parallel }(1-t)x+ty{\mathrm{\parallel }}_2\le (1-t)\mathrm{\parallel }x{\mathrm{\parallel }}_2+t\mathrm{\parallel }y{\mathrm{\parallel }}_2,$ so the norm is convex.

If $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le 1$, [L4] gives $⟨v,y⟩\le \mathrm{\parallel }y{\mathrm{\parallel }}_2$, so $v$ is a subgradient of the norm at zero. Conversely, a subgradient $v$ satisfies $\mathrm{\parallel }y{\mathrm{\parallel }}_2\ge ⟨v,y⟩$ for all $y$; taking $y=v$ gives $\mathrm{\parallel }v{\mathrm{\parallel }}_2\ge \mathrm{\parallel }v{\mathrm{\parallel }}_2^2$, hence $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le 1$.

Since $q(x)={\sum }_{j<n}{x}_j^2$, coordinatewise differentiation with [L5]–[L7] gives gradient $2x$ and constant Hessian $2I$, so $q$ is $C^2$ and its Hessian is positive definite. Thus [L2] gives convexity, [L3] gives strict convexity, and differentiable subgradient uniqueness gives $\mathrm{\partial }q(x)=\{2x\}$.

Each affine constituent is convex, so [L1] makes their nonempty finite maximum $f$ convex.

If $j\in J(x)$, then $f(y)\ge ⟨a_j,y⟩+b_j=f(x)+⟨a_j,y-x⟩$. Nonnegative weighted sums of these inequalities show that every convex combination of active slopes is a subgradient.

The active-weight simplex is closed and bounded, hence compact by [L3]; its affine image is compact by [L4] and closed by [L5]. If a subgradient $v$ lay outside this active convex hull, [L2] would give a direction $h$ with $⟨v,h⟩$ strictly larger than every active $⟨a_j,h⟩$. Finiteness lets one choose $t>0$ small enough that inactive affine pieces remain below the active maximum at $x+th$. Then the subgradient inequality would require an increment at least $t⟨v,h⟩$, while the actual increment is $t{\max }_{j\in J(x)}⟨a_j,h⟩$, a contradiction. The displayed four-piece formula follows from its four active slopes at zero.

A convex combination of two domain points equals zero only when every endpoint having positive weight is zero. In that case [F1] is an equality. Otherwise the left side is zero and the right side is nonnegative, so [F1] again holds.

For every positive $x$, $f(x)=0$, whereas $f(0)=1$, so the right-hand limit at zero is not the function value. The function is convex on $[0,1]$ but is not continuous at $0$. This does not contradict [L1], because the domain is not open at zero.

Put $s_x=\sqrt{1-\mathrm{\parallel }x{\mathrm{\parallel }}_2^2}$. The lifted vectors $(x,s_x)$ have norm one in ${\mathbb{R}}^{n+1}$. By the triangle inequality in [L2], the convex combination $((1-t)x+ty,(1-t)s_x+t{s}_y)$ has norm at most one, so its nonnegative last coordinate is at most $s_{(1-t)x+ty}$. Negating gives the convexity inequality for $f$ on the whole closed ball.

Fix a unit vector $u$ and suppose $v$ satisfied [F1] at $u$. Testing $y=\alpha u$ for $0\le \alpha <1$ gives $⟨v,u⟩\ge \frac{\sqrt{1-{\alpha }^2}}{1-\alpha }=\sqrt{\frac{1+\alpha }{1-\alpha }}\mathrm{.}$ The right side is unbounded as $\alpha$ approaches $1$, impossible for the fixed vector $v$. Hence $\mathrm{\partial }f(u)=\mathrm{\varnothing }$.

Step 2.1 applies to every sphere point, while [L1] gives nonempty subdifferentials at every interior point. Thus the interior hypothesis in the existence theorem is sharp.

The weak sublevel $\{g\le \alpha \}$ is empty for $\alpha <-1$, equals $\{0,1\}$ at $\alpha =-1$, equals $\{0\}\cup [-\alpha ,1]$ for $-1<\alpha <0$, and is all of $[0,1]$ for $\alpha \ge 0$. Each is relatively closed, so [L1] makes $g$ lower semicontinuous.

Every value of $g$ is negative, but $g(1\mathrm{/}N)=-1\mathrm{/}N$ approaches zero through positive natural $N$. Hence $\sup g([0,1])=0$ and the supremum is not attained, so $g$ has no maximum.

The function does attain its minimum $-1$ at zero, as [L2] requires. It is only the unsupported opposite extremum that fails.

Strict superlevels of $1_E$ are empty, $E$, or all of $X$, while strict sublevels are empty, $X\setminus E$, or all of $X$. By [L1] and [L2], $1_E$ is lower semicontinuous when $E$ is open and upper semicontinuous when $E$ is closed.

For $E=\{0\}$ in $[0,1]$, step 1.1 gives upper semicontinuity, but values at positive points tend to zero rather than the value one at zero. Negation exchanges upper and lower semicontinuity, so $-1_E$ supplies the lower-semicontinuous discontinuous example.

By [L2] and [L3], the Hessian of $f(x,y)=x^2$ is $\mathrm{diag}(2,0)$, which is positive semidefinite. By [L1], $f$ is convex.

The distinct points $(0,0)$ and $(0,1)$, and every point of the segment joining them, all have value zero. Thus the strict convexity inequality is an equality on that segment. The Hessian of $f(x,y)=x^2$ is positive semidefinite everywhere, but $f$ is not strictly convex.

Let $r<s$ and $0<t<1$, and put $c=(1-t)r+ts$. Apply [L2] on $[r,c]$ and $[c,s]$. The two secant slopes equal $4u^3$ and $4v^3$ for some $r<u<c<v<s$ by [L1], so the first is strictly smaller; rearranging gives $c^4<(1-t)r^4+t{s}^4$. Thus $x^4$ is strictly convex. For distinct vectors, at least one coordinate gives this strict inequality and every other coordinate gives the corresponding weak one; adding shows that $f(x)={\sum }_{j<n}{x}_j^4$ is strictly convex for $n\ge 1$.

Differentiating again with [L1] and [L3], the Hessian is diagonal with entries $12x_j^2$ and zero off-diagonal entries. At the origin it is the zero matrix, hence singular and not positive definite because $n\ge 1$. Thus $f(x)={\sum }_{j<n}{x}_j^4$ is strictly convex, but its Hessian at the origin is the zero matrix.

The interval in [L1] is convex and the displayed function is convex on it.

The same function fails continuity at the boundary point zero by [L1], so it refutes the claim. Openness of the domain, or restriction to its interior, is essential.

In positive dimension, the closed unit ball in [L1] contains more than one vector, so the norm has a nonsingleton subdifferential at zero.

By [L2], differentiability there would force the subdifferential to be a singleton. Hence the convex Euclidean norm is not differentiable at zero, and the claim is false.

The first function in [L1] satisfies upper semicontinuity on a compact domain and fails continuity.

Its negative in [L1] separately satisfies lower semicontinuity and fails continuity on the same compact domain. Thus neither one-sided notion implies continuity.

The function in [L1] satisfies the Hessian hypothesis of the false claim.

It is constant along every vertical line and therefore violates strict convexity by [L1]. Hence the claim is false.

The function in [L1] satisfies strict convexity globally.

Its Hessian at the origin is zero and hence not positive definite by [L1], directly refuting the claim.