Intuitively, strong convexity is convexity with “actual curvature”. (Recall that convex function may not be curved at all: e.g., linear functions.)
<aside> 💡 Definition: Strong convexity
A function $f$ is $\alpha$-strongly convex (for $\alpha \geq 0$) over a convex and closed $S \subseteq \mathrm{dom} f$ if for any $x \in S$, there exists $g_x \in \partial f(x)$ such that:
$$ \forall ~ y\in S,\qquad f(y) \geq f(x)+ g_x \cdot (y-x) + \frac\alpha2 \|y-x\|^2. $$
In particular, a differentiable $f$ is $\alpha$-strongly convex over $S$ if for any $x \in S$,
$$ \forall ~ y\in S,\qquad f(y) \geq f(x)+ \nabla f(x) \cdot (y-x) + \frac\alpha2 \|y-x\|^2. $$
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In words: a function $f$ is strongly convex if at any given point $x$ in its domain, we can lower bound the function with a quadratic tangential to $f$ at $x$.
Contrast this with smoothness, which imposes an analogous upper bound.
This is a generalization of the first-order characterization of convexity, which is recovered when $\alpha=0$. A function satisfying the condition with $\alpha=0$ is said to be convex, but not strongly convex.
There is also a “zero-order” characterization for $\alpha$-strong convexity. We omit it here since it will not be very useful. But note the definition above does not require differentiability!
Strong convexity implies a tangential quadratic lower bound (in red) at every point in the domain.