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Suppose $f$ is twice continuously differentiable convex for $x \in \mathbb{R}$, then $f''(x) \succeq 0$.

Proof: Denote $x_\tau=x+\tau s, \tau > 0$. Then $0 \leq \frac{1}{\tau} \langle f'(x_\tau) - f'(x), x_\tau - x\rangle=\frac{1}{\tau} \langle f'(x_\tau) - f'(x),s\rangle =\frac{1}{\tau}\int_0^\tau \langle f''(x+\lambda s)s, s\rangle d\lambda$.

We get $f''(x) \succeq 0$ by tending $\tau$ to zero.

My questions about this proof follow. First, I thought the correct expressions should be $0 \leq \frac{1}{\tau} \langle f'(x_\tau) - f'(x), x_\tau - x\rangle=\langle f'(x_\tau) - f'(x),s\rangle =\int_0^\tau \langle f''(x+\lambda s)s, s\rangle d\lambda$.

Second, how do I get $f''(x) \succeq 0$ by tending $\tau$ to zero? If we let $\tau$ tend to zero, the integral also tends to zero.

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You are right, that the correct expressions should be $$\frac{1}{\tau} \langle f'(x_\tau) - f'(x), x_\tau - x\rangle=\langle f'(x_\tau) - f'(x),s\rangle.$$ There must be a typo in the textbook.

Second you seem to have to some confusion about the domain of $f$. If the domain of $f$, i.e. $x \in \mathbb{R}$ then $f''(x) \in \mathbb{R}$ so the $\succeq$ symbol does not make sense you can just use normal $\ge$ symbol .

The next question you are asking is that why does $$\begin{align}s^2 \int_0^\tau f''(x+\lambda s) d\lambda \ge 0, \qquad \forall \tau > 0 \qquad (1)\end{align}$$ imply that $f''(x) \ge 0$.

This is where the "continuously differentiable" condition on $f''$ becomes important. If $f''(x)$ was smaller than $0$ then you could find a small neighborhood around $x$ where it will still be smaller than $0$ because it is continuous. Then you could take $\tau$ to be small enough so that it lies in that neighborhood, and then (1) will be false, therefore creating a contradiction.

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