A function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is convex if for all $x,y\in \mathbb{R}^n$ the inequality $$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$$ holds for all $t\in [0,1]$.
I want to show that if a twice continuously differentiable funtion $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is convex then the Hessian matrix is positive semidefinite for all $x\in \mathbb{R}^n$.
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I have done the following :
Let $g(t)=f(\mathbf{x}+t\mathbf{v})$. The first and second derivatives of $f$ are
\begin{align*}g'(t) &= \frac{d}{d{t}}f( \mathbf{x} + t \mathbf{v})=\sum_i\frac{\partial}{\partial{f_i}}f( \mathbf{x} + t \mathbf{v})\cdot \frac{d{f_i}}{d{t}}\\ & =\sum_i\frac{\partial}{\partial{f_i}}f( \mathbf{x} + t \mathbf{v})\cdot \mathbf{v}
\\ g''(t) &= \frac{d}{d{t}}\left (\sum_i\frac{\partial}{\partial{f_i}}f(\mathbf{x} + t \mathbf{v})\cdot \mathbf{v}\right )\\ &=\left (\sum_i\frac{d}{d{t}}\left[\frac{\partial}{\partial{f_i}}f(\mathbf{x} + t \mathbf{v})\right]\right )\cdot \mathbf{v}\\ &=\left (\sum_i\sum_j\frac{\partial}{\partial{f_j}}\frac{\partial}{\partial{f_i}}f(\mathbf{x} + t \mathbf{v})\cdot \frac{d f_j}{d t}\right )\cdot \mathbf{v} \\ & =
\left (\sum_i\sum_j\frac{\partial}{\partial{f_j}}\frac{\partial}{\partial{f_i}}f(\mathbf{x} + t \mathbf{v})\cdot \mathbf{v}\right )\cdot \mathbf{v} \\ & =\left (\sum_i\sum_j\frac{\partial}{\partial{f_j}}\frac{\partial}{\partial{f_i}}f(\mathbf{x} + t \mathbf{v})\right )\cdot \mathbf{v}\cdot \mathbf{v}\\ & =\mathbf{v}^T\cdot \left (\sum_i\sum_j\frac{\partial}{\partial{f_j}}\frac{\partial}{\partial{f_i}}f(\mathbf{x} + t \mathbf{v})\right )\cdot \mathbf{v}
\end{align*}
We assume that the Hessian matrixis not positive semidefinit, so there is a vector $\mathbf{v}$ such that $\mathbf{v}^T\cdot \left (\sum_i\sum_j\frac{\partial}{\partial{f_j}}\frac{\partial}{\partial{f_i}}f(\mathbf{x} + t \mathbf{v})\right )\cdot \mathbf{v}<0$,i.e. $g''(t)<0$.
Since $f$ is convex,we have that $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$.
How do we get a contradiction? Or do we use another way to get the desired result?
Best Answer
We define $g(t)=f(x+tv)$, let $\lambda \in (0,1)$
\begin{align} g(\lambda t_1 + (1-\lambda) t_2) &= f(x + (\lambda t_1 + (1-\lambda) t_2)v)\\ &=f(\lambda (x+t_1v) + (1-\lambda)(x + t_2v))\\ &\le \lambda f(x + t_1v) +(1-\lambda)f(x+t_2v)\\ &=\lambda g(t_1) + (1-\lambda) g(t_2) \end{align}
Hence, $g$ is convex.