$\mu$-strongly convex is strictly convex

Question

From this definition, how to prove $\mu$-strongly convex is strictly convex? I know there are some similar questions in the website, but they use the definition containing differentiable. But that is not necessary.

The defition of $\mu$-strongly convex is:

f is $\mu$-strongly convex on T (with respect to the Euclidean norm $||\cdot ||$) if there exists a $\mu$ > 0 such that for any $\theta, \tau \in X$ and $\lambda \in [0, 1]$ we have

$$f(\lambda \theta + (1-\lambda) \tau) \leq \lambda f(\theta) + (1-\lambda) f(\tau) + \mu \lambda (1-\lambda) || \theta - \tau||^2$$

I know if I can drop the last term above, then it is done. But how to justify this drop step? From the definition, it only says there exists a $\mu$, it doesn't say any $\mu >0$, so I cannot say let $\mu \to 0$ and drop the last term.

Answer 1

@Arun G is correct: the correct definition of strongly convex is

Definiton. A function $f \colon (X, \| \cdot \|) \to \mathbb R$ is $m$-strongly convex if there exists a $m > 0$ such that $$ f\big( (1 - t) x + t y \big) \le (1 - t) f(x) + t f(y) - \frac{1}{2} m t (1 - t) \| x - y \|^2 \qquad \forall x, y \in X \ \forall t \in (0, 1). $$

(Note that the inequality is always an equality for $t \in \{ 0, 1 \}$, so we can restrict ourselves to $t \in (0, 1)$.

Now let $x, y \in X$ with $x \ne y$ and $t \in (0, 1)$. If $f$ is $m$-strongly convex for some $m > 0$, then $$ f\big( (1 - t) x + t y \big) \le (1 - t) f(x) + t f(y) - \frac{1}{2} m t (1 - t) \| x - y \|^2. $$ Since $x \ne y$, we have $\| x - y \| > 0$ and thus (since $m, t, 1 - t > 0$) $$ f\big( (1 - t) x + t y \big) < (1 - t) f(x) + t f(y), $$ that is, $f$ is strictly convex.