So i have to prove that the inequality below is true if and only if f is convex and Lipschitz continuous. i have the first part down which is to assume f is convex and show the inequality. But i cant seem to figure out the second part of the proof which is to show that f is convex given the inequality.
So my question is how to prove f is convex given, $$0 \le f(y) - f(x) \le \langle \nabla f(x),y-x\rangle \le \frac{L}{2}||x-y||^2$$
I am having trouble with the derivative term. I think i might be able to use the FTC to convert it to a limit but am stuck.
I think there is something wrong with your inequalities. According to what you're writing, it should be $$ f(y) \leq f(x) + \langle \nabla f(x), y-x\rangle, $$ but instead if $f$ is convex then for every $x,y$ it is $$ f(y) \geq f(x) + \langle \nabla f(x), y-x\rangle, $$ that is, the graph of $f$ lies above its tangent line at a point $x$. This is often used as the definition of convexity (of continuously diff/ble functions) and it is equivalent to the other well-known definition: A function $f$ is convex if for all $\tau\in [0,1]$ and all $x,y$, we have $f(\tau x+(1-\tau) y) \leq \tau f(x) + (1-\tau)f(y)$.
Let us denote, for convenience, $x_\tau = \tau x + (1-\tau) y$. Then, assuming that $f(y) \geq f(x) + \langle \nabla f(x), y-x\rangle$ for all $x,y$ we have
$$ f(x_\tau) \leq f(y) + \langle \nabla f(x_\tau), y-x_\tau \rangle = f(y) + \tau \langle \nabla f(x_\tau), y-x\rangle $$ and $$ f(x_\tau) \leq f(x) + \langle \nabla f(x_\tau), x-x_\tau \rangle = f(x) - (1-\tau) \langle \nabla f(x_\tau), y-x\rangle $$ We now multiply the first inequality by $1-\tau$ and the second one by $\tau$ and we add them by parts to get $f(\tau x+(1-\tau) y) \leq \tau f(x) + (1-\tau)f(y)$.
Let us now assume that $f(\tau x+(1-\tau) y) \leq \tau f(x) + (1-\tau)f(y)$ for all $x,y$ and $\tau\in [0,1]$. Then
$$ \begin{align*} f(y) &\geq \tfrac{1}{1-\tau}[f(x_\tau) - \tau f(x) ] = f(x) + \tfrac{1}{1-\tau}[f(x_\tau) - f(x)]\\ &= f(x) + \tfrac{1}{1-\tau}[f(x + (1-\tau)(y-x))-f(x)], \end{align*} $$ and letting $\tau\to 1$, that is, $1-\tau \to 0$ we have (by the definition of directional derivative) $$ \lim_{h\to 0}\tfrac{1}{h}[f(x + h(y-x))-f(x)] = \langle \nabla f(x), y-x\rangle $$ which completes the proof. $\square$