Why does $u^T H u $ give the second derivatives in the $u$ direction? I would like to have an intuition behind the fact that a positive definite Hessian is equivalent to the fact that the function is convex.
In fact, given a function $f : \mathbb{R}^2 \to \mathbb{R}$, if we put $\boldsymbol{u} = \begin{bmatrix}x ,&y\end{bmatrix}$, $\mathbf{H}_f= \begin{bmatrix} f_{xx} &f_{xy}\\f_{yx} &f_{yy}\end{bmatrix}$ by developing $\boldsymbol{u}^\intercal \mathbf{H}_f \boldsymbol{u}$ we obtain a quadratic form.
I don't understand how this quadratic form gives the value of the second derivative of $f$ in the direction of $\boldsymbol{u}$.
Thank you for your attention!
 A: I don't know that there's an easier way to see it than by direct computation:
\begin{align*}
\lim_{t\to 0} \frac{d^2}{dt^2} f(\mathbf{x}+t\mathbf{u}) &= \lim_{t\to 0} \frac{d}{dt}\mathbf{u}^T\nabla f(\mathbf{x} + t\mathbf{u})\\
& = \lim_{t\to 0} \mathbf{u}^T Hf(\mathbf{x}+t\mathbf{u})\mathbf{u}\\
& = \mathbf{u}^T Hf(\mathbf{x})\mathbf{u}.
\end{align*}
What's really going on here is that the Hessian is the second-order part of the Taylor expansion of a multivariable function; it plays the multivariable role of the second derivative.
$$f(\mathbf{x}+t\mathbf{u}) = f(\mathbf{x}) + t \mathbf{u}^T\nabla f(\mathbf{x}) + \frac{t^2}{2} \mathbf{u}^T Hf(\mathbf{x}) \mathbf{u} + \frac{t^3}{3!} \sum_{ijk}\frac{\partial^3 f}{\partial x_i\partial x_j\partial x_k}(\mathbf{x})u_iu_ju_k + \cdots$$
though I was flabbergasted a few years ago to learn that Taylor's theorem isn't even mentioned once in the entire 500-page multivariable calculus textbook popular at my university, so it's perhaps no surprise if you haven't seen it...
