Proof of Properties of the Subdifferential on $\mathbb{R}^n$: Minimisers, Convexity and Closedness, Sums and Positive Multiples

Throughout, membership in a subdifferential is read off from Subgradients and the Subdifferential of a Real-Valued Function on a Subset of $\mathbb{R}^n$ (ii) and Subgradients and the Subdifferential of a Real-Valued Function on a Subset of $\mathbb{R}^n$ (i): $a\in\partial f(x)$ means

$$f(y)\ge f(x)+a\cdot(y-x)\qquad\text{for every }y\in C,$$

and the bilinearity of the dot product and $0\cdot v=0$ are background facts of the setting.

For (i), the subgradient inequality for $a=0$ reads $f(y)\ge f(x)+0\cdot(y-x)=f(x)$ for every $y\in C$; so $0\in\partial f(x)$ exactly when $x$ minimises $f$ on $C$.

For (ii), let $a,a'\in\partial f(x)$, $s\in[0,1]$ and $y\in C$. By bilinearity and $1-s\ge0$, $s\ge0$,

$$\bigl((1-s)a+sa'\bigr)\cdot(y-x)=(1-s)\,a\cdot(y-x)+s\,a'\cdot(y-x)\le(1-s)\bigl(f(y)-f(x)\bigr)+s\bigl(f(y)-f(x)\bigr)=f(y)-f(x).$$

So $(1-s)a+sa'\in\partial f(x)$, which is convexity of $\partial f(x)$ by Segments and Convex Subsets of $\mathbb{R}^n$ (i) and Segments and Convex Subsets of $\mathbb{R}^n$ (ii).

For (iii), by Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (iv) and Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (i) it suffices to show $\operatorname{cl}\partial f(x)\subseteq\partial f(x)$. Let $a\in\operatorname{cl}\partial f(x)$ and $y\in C$, and let $\varepsilon>0$ be given. By Closure of a Subset of a Metric Space (i) there is $a'\in\partial f(x)$, depending on $\varepsilon$, with $d_{E}(a,a')<\varepsilon$, that is, $\lVert a'-a\rVert<\varepsilon$ (Open Ball in a Metric Space (ii), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii)). By The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ (i),

$$f(y)-f(x)\ge a'\cdot(y-x)=a\cdot(y-x)+(a'-a)\cdot(y-x)\ge a\cdot(y-x)-\varepsilon\lVert y-x\rVert .$$

Since $\varepsilon>0$ was arbitrary, $f(y)-f(x)\ge a\cdot(y-x)$; so $a\in\partial f(x)$.

For (iv), let $a\in\partial f(x)$, $a'\in\partial g(x)$ and $y\in C$. Adding the two subgradient inequalities gives

$$f(y)+g(y)\ge f(x)+g(x)+(a+a')\cdot(y-x),$$

so $a+a'\in\partial(f+g)(x)$.

For (v), let $a\in\mathbb{R}^{n}$ and $y\in C$. Since $\lambda>0$,

$$\lambda f(y)\ge\lambda f(x)+a\cdot(y-x)\iff f(y)\ge f(x)+\bigl(\lambda^{-1}a\bigr)\cdot(y-x).$$

So $a\in\partial(\lambda f)(x)$ if and only if $\lambda^{-1}a\in\partial f(x)$. If this holds, then $a=\lambda a''$ with $a''=\lambda^{-1}a\in\partial f(x)$; conversely, if $a=\lambda a''$ with $a''\in\partial f(x)$, then $\lambda^{-1}a=a''\in\partial f(x)$. Hence $a\in\partial(\lambda f)(x)$ if and only if $a=\lambda a''$ for some $a''\in\partial f(x)$.