Lemma 21.11.63.66B643EAC79A Let $I\subseteq \mathbb {R}$ be an open interval containing $0$. Let $(f_ k\colon I\rightarrow \mathbb {R})_{k\geq 1}$ be a sequence of convex functions, and let $g\colon I\to \mathbb {R}$ be a function. Assume that
\[ \varlimsup _{k\to \infty } f_ k(t)\leq g(t)\quad \text{for every } t\in I, \]
and that
\[ \lim _{k\to \infty } f_ k(0)=g(0). \]
If each $f_ k$ and $g$ is differentiable at $0$, then
\[ \lim _{k\to \infty } f_ k'(0)=g'(0). \]
Proof.
Since $f_ k$ is convex and differentiable at $0$, for every $t\in I$,
\[ f_ k(t)\geq f_ k(0)+t f_ k'(0). \]
Taking the upper limit as $k\to \infty $ and using the assumptions gives
\[ \varlimsup _{k\to \infty } t f_ k'(0) \leq g(t)-g(0). \]
For $t>0$, this yields
\[ \varlimsup _{k\to \infty } f_ k'(0) \leq \frac{g(t)-g(0)}{t}. \]
Letting $t\to 0+$ and using the differentiability of $g$ at $0$, we get
\[ \varlimsup _{k\to \infty } f_ k'(0)\leq g'(0). \]
For $t<0$, the same inequality gives
\[ \varliminf _{k\to \infty } f_ k'(0) \geq \frac{g(t)-g(0)}{t}. \]
Letting $t\to 0-$ and again using the differentiability of $g$ at $0$, we obtain
\[ \varliminf _{k\to \infty } f_ k'(0)\geq g'(0). \]
Therefore
\[ \lim _{k\to \infty } f_ k'(0)=g'(0), \]
as claimed.
$\square$
Pome
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