How does $f_0(z(0))' < 0 \Rightarrow f_0(z(t)) \lt f_0(x)$?

Question

From Convex optimization by Boyd and Vanderberghe:

In the red box in the proof below:

How does the fact that the derivative of $f_0(z(t))$ at $t=0$ show that $f_0(z(t)) \lt f_0(x)$?

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Answer 1

Consider the function $g(t) = f_0(z(t))$. It is given that $$ g'(0) = \lim_{t \to 0} \frac{g(t) - g(0)}{t-0} $$ exists and is negative. From the definition of the limit it follows that there is a $\delta > 0$ such that $$ \frac{g(t) - g(0)}{t-0} < \frac 12 g'(0) $$ for $|t| < \delta$. Consequently, for $0 < t < \delta$, $$ g(t) < g(0) + \frac t2 g'(0) < g(0) \, . $$