Theorem: Continuity of Convex Real Function at Interior Point implies Local Lipschitz Continuity

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\GF$.
Let $C \subseteq X$ be an open convex set with non-empty interior $\Int C$. 
Let $x \in \Int C$. 
Let $f : C \to \R$ be a convex function such that:
:$f$ is continuous at $x$.
Let $\map {B_\delta} x$ denote the open ball centered at $x$ with radius $\delta$.
Then there exists $\delta > 0$ such that $f$ is Lipschitz continuous on $\map {B_\delta} x \subseteq C$.

Proof:
Since $f$ is continuous at $x$, there exists $\delta > 0$ such that:
:$\norm {u - x}_X < 2 \delta$
implies that:
:$\size {\map f u - \map f x} < \dfrac 1 2$
for $u \in C$. 
Then if:
:$\norm {w - x}_X < 2 \delta$
and:
:$\norm {z - x}_X < 2 \delta$
we have:
:$\size {\map f w - \map f z} < 1$
by the Triangle Inequality.
Take $\delta$ sufficiently small to have $\map {B_{2 \delta} } x \subseteq C$. 
Let $y, z \in \map {B_\delta} x$ be distinct. 
Let:
:$\alpha = \norm {y - z}_X \ne 0$
Let:
:$\ds w = y + \frac \delta \alpha \paren {y - z}$
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {w - y}_X
		| r = \frac \delta \alpha \norm {y - z}_X
		| c = {{NormAxiomVector|2}}
}}
{{eqn	| r = \frac \delta \alpha \times \alpha
}}
{{eqn	| r = \delta 
}}
{{end-eqn}}
Since $y \in \map {B_\delta} x$, we have:
{{begin-eqn}}
{{eqn	| l = \norm {w - x}_X
		| o = \le
		| r = \norm {w - y}_X + \norm {y - x}_X 
}}
{{eqn	| o = <
		| r = \delta + \delta 
}}
{{eqn	| r = 2 \delta 
}}
{{end-eqn}}
and so $w \in \map {B_{2 \delta} } x$.
Rearranging the equation:
:$\ds w = y + \frac \delta \alpha \paren {y - z}$
we obtain:
:$\ds \paren {1 + \frac \delta \alpha} y = w + \frac \delta \alpha z$
Multiplying by:
:$\ds \frac 1 {1 + \frac \delta \alpha}$
we have:
:$\ds y = \frac 1 {1 + \frac \delta \alpha} w + \frac {\frac \delta \alpha} {1 + \frac \delta \alpha} z$ 
We have:
:$\ds \frac 1 {1 + \frac \delta \alpha} = \frac \alpha {\alpha + \delta}$
and:
:$\ds \frac {\frac \delta \alpha} {1 + \frac \delta \alpha} = \frac \delta {\alpha + \delta}$
so that:
:$\ds y = \frac \alpha {\alpha + \delta} w + \frac \delta {\alpha + \delta} z$ with $w, z \in \map {B_{2 \delta} } x \subseteq C$. 
Since:
:$\ds \frac \alpha {\alpha + \delta} + \frac \delta {\alpha + \delta} = 1$
we have:
:$\ds \map f y \le \frac \alpha {\alpha + \delta} \map f w + \frac \delta {\alpha + \delta} \map f z$
from the convexity of $f$.
We then have:
{{begin-eqn}}
{{eqn	| l = \map f y - \map f z
		| o = \le 
		| r = \frac \alpha {\alpha + \delta} \map f w - \paren {1 - \frac \delta {\alpha + \delta} } \map f z
}}
{{eqn	| r = \frac \alpha {\alpha + \delta} \map f w - \frac \alpha {\alpha + \delta} \map f z
}}
{{eqn	| r = \frac \alpha {\alpha + \delta} \paren {\map f w - \map f z}
}}
{{end-eqn}}
Since $z \in \map {B_{2 \delta} } x$ and $w \in \map {B_{2 \delta} } x$, we have that:
:$\size {\map f w - \map f z} < 1$
and in particular:
:$\map f w - \map f z < 1$
Hence:
:$\ds \map f y - \map f z < \frac \alpha {\alpha + \delta}$
Since $y, z \in \map {B_\delta} x$ are arbitrary and distinct, we also have:
:$\ds \map f z - \map f y < \frac \alpha {\alpha + \delta}$
Hence:
{{begin-eqn}}
{{eqn	| l = \size {\map f y - \map f z}
		| o = <
		| r = \frac \alpha {\alpha + \delta} 
}}
{{eqn	| o = <
		| r = \frac \alpha \delta 
}}
{{eqn	| r = \frac 1 \delta \norm {y - z}_X 
}}
{{end-eqn}}
for $y, z \in \map {B_\delta} x$. 
Hence $f$ is Lipschitz continuous on $\map {B_\delta} x$ with Lipschitz constant $1/\delta$. 
{{qed}}