[proofplan]
The forward direction restricts the convex function $f$ to the line segment from $x$ to $y$ and uses the difference quotients at the initial point of the segment. Convexity gives an upper bound on these quotients in terms of $f(y)-f(x)$, and differentiability identifies their limit with $\nabla f(x)\cdot(y-x)$. For the reverse direction, we apply the assumed supporting-hyperplane inequality at the midpoint-like point $z=(1-\lambda)x+\lambda y$, once with target $x$ and once with target $y$. The two gradient terms cancel after multiplying by the convex weights, leaving exactly the convexity inequality.
[/proofplan]
[step:Derive the supporting inequality from convexity along a line segment]
Assume first that $f$ is convex on $U$. Fix $x,y \in U$. Since $U$ is convex, the map
\begin{align*}
\gamma_{x,y}: [0,1] \to U, \qquad t \mapsto x+t(y-x)
\end{align*}
is well-defined. Define the real-valued function
\begin{align*}
\varphi_{x,y}: [0,1] \to \mathbb{R}, \qquad t \mapsto f(\gamma_{x,y}(t)).
\end{align*}
For every $t \in (0,1]$, convexity of $f$ applied to $\gamma_{x,y}(t)=(1-t)x+ty$ gives
\begin{align*}
\varphi_{x,y}(t) \leq (1-t)\varphi_{x,y}(0)+t\varphi_{x,y}(1).
\end{align*}
Rearranging this inequality and dividing by $t>0$ gives
\begin{align*}
\frac{\varphi_{x,y}(t)-\varphi_{x,y}(0)}{t} \leq \varphi_{x,y}(1)-\varphi_{x,y}(0).
\end{align*}
Because $f$ is differentiable at $x$, the directional derivative of $f$ at $x$ in the direction $y-x$ exists and equals $\nabla f(x)\cdot(y-x)$. Equivalently,
\begin{align*}
\lim_{t\downarrow 0}\frac{f(x+t(y-x))-f(x)}{t}=\nabla f(x)\cdot(y-x).
\end{align*}
Taking the limit as $t\downarrow 0$ in the previous inequality therefore yields
\begin{align*}
\nabla f(x)\cdot(y-x) \leq f(y)-f(x).
\end{align*}
This is precisely
\begin{align*}
f(y) \geq f(x)+\nabla f(x)\cdot(y-x).
\end{align*}
[guided]
Assume that $f$ is convex on $U$, and fix two points $x,y \in U$. The desired inequality compares $f(y)$ with the affine approximation to $f$ at $x$, so we examine $f$ only along the line segment beginning at $x$ and ending at $y$. Since $U$ is convex, every point of this segment lies in $U$, so the map
\begin{align*}
\gamma_{x,y}: [0,1] \to U, \qquad t \mapsto x+t(y-x)
\end{align*}
is well-defined. Define
\begin{align*}
\varphi_{x,y}: [0,1] \to \mathbb{R}, \qquad t \mapsto f(\gamma_{x,y}(t)).
\end{align*}
The point $\gamma_{x,y}(t)$ is the convex combination $(1-t)x+ty$. Therefore convexity of $f$ gives, for each $t \in (0,1]$,
\begin{align*}
f(x+t(y-x)) \leq (1-t)f(x)+tf(y).
\end{align*}
In terms of $\varphi_{x,y}$, this is
\begin{align*}
\varphi_{x,y}(t) \leq (1-t)\varphi_{x,y}(0)+t\varphi_{x,y}(1).
\end{align*}
We now isolate a difference quotient at the point $t=0$. Subtracting $\varphi_{x,y}(0)$ from both sides gives
\begin{align*}
\varphi_{x,y}(t)-\varphi_{x,y}(0) \leq t(\varphi_{x,y}(1)-\varphi_{x,y}(0)).
\end{align*}
Since $t>0$, division by $t$ preserves the inequality:
\begin{align*}
\frac{\varphi_{x,y}(t)-\varphi_{x,y}(0)}{t} \leq \varphi_{x,y}(1)-\varphi_{x,y}(0).
\end{align*}
The left-hand side is the one-sided difference quotient of $f$ at $x$ in the direction $y-x$. Because $f$ is differentiable at $x$, its first-order expansion at $x$ implies
\begin{align*}
\lim_{t\downarrow 0}\frac{f(x+t(y-x))-f(x)}{t}=\nabla f(x)\cdot(y-x).
\end{align*}
The right-hand side is independent of $t$, so taking $t\downarrow 0$ gives
\begin{align*}
\nabla f(x)\cdot(y-x) \leq f(y)-f(x).
\end{align*}
Rearranging gives the supporting inequality
\begin{align*}
f(y) \geq f(x)+\nabla f(x)\cdot(y-x).
\end{align*}
[/guided]
[/step]
[step:Use the supporting inequality at an intermediate point]
Conversely, assume that for every $a,b \in U$,
\begin{align*}
f(b) \geq f(a)+\nabla f(a)\cdot(b-a).
\end{align*}
Fix $x,y \in U$ and $\lambda \in [0,1]$. If $\lambda=0$ or $\lambda=1$, the convexity inequality is an equality. Hence assume $\lambda \in (0,1)$.
Define the intermediate point
\begin{align*}
z := (1-\lambda)x+\lambda y.
\end{align*}
Since $U$ is convex, $z \in U$. Applying the assumed inequality with $a=z$ and $b=x$ gives
\begin{align*}
f(x) \geq f(z)+\nabla f(z)\cdot(x-z).
\end{align*}
Applying it with $a=z$ and $b=y$ gives
\begin{align*}
f(y) \geq f(z)+\nabla f(z)\cdot(y-z).
\end{align*}
Multiply the [first inequality](/theorems/2897) by $1-\lambda$ and the second by $\lambda$, then add:
\begin{align*}
(1-\lambda)f(x)+\lambda f(y) \geq f(z)+\nabla f(z)\cdot((1-\lambda)(x-z)+\lambda(y-z)).
\end{align*}
By the definition of $z$,
\begin{align*}
(1-\lambda)(x-z)+\lambda(y-z)=(1-\lambda)x+\lambda y-z=0.
\end{align*}
Therefore
\begin{align*}
(1-\lambda)f(x)+\lambda f(y) \geq f(z).
\end{align*}
Substituting $z=(1-\lambda)x+\lambda y$ gives
\begin{align*}
f((1-\lambda)x+\lambda y) \leq (1-\lambda)f(x)+\lambda f(y).
\end{align*}
Since $x,y \in U$ and $\lambda \in [0,1]$ were arbitrary, $f$ is convex on $U$.
[/step]
