Upper bound: Given $L$-smooth convex $f$; $( y- x)^T \left( \nabla f(z)-\nabla f(x)\right)\leq(L/2) ( \| x-z\|^2+\| x-y\|^2+\| z-y\|^2)$?

Question

Given $L$-smooth convex $f$, I would highly appreciate if you can confirm whether the following bound is correct or not.

\begin{align} \left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right) \leq \frac{L}{2} \left( \| x-z\|^2+\| x-y\|^2+\| z-y\|^2 \right) \tag{$\clubsuit$}. \end{align}

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Attempt:

Since function $f$ is both $L$-smooth and convex, I particularly make use of the following two inequalities (e.g., can be found here or many other books such as in Yurii Nesterov's book).

1. Three points descent lemma: \begin{align} f(x) \leq f(y) + \left( x - y \right)^T \nabla f(z) + \frac{L}{2} \| x - z \|^2 \end{align}
2. $$0 \leq f(y) - f(x) - \left( y - x\right)^T \nabla f(x) \leq \frac{L}{2} \| x - y \|^2 $$

To this end, we rewrite the above two respective inequalities

1. \begin{align} f(x) &\leq f(y) + \left( x - y \right)^T \nabla f(z) + \frac{L}{2} \| x - z \|^2 \\ \Longleftrightarrow \left( y - x \right)^T \nabla f(z) &\leq f(y) - f(x) + \frac{L}{2} \| x - z \|^2 \tag{1} \end{align} 2. \begin{align} f(y) - f(x) - \left( y - x\right)^T \nabla f(x) &\leq \frac{L}{2} \| x - y \|^2 \\ \Longleftrightarrow \left( y - x\right)^T \nabla f(x) &\geq f(y) - f(x) - \frac{L}{2} \| x - y \|^2 \tag{2} \end{align}

Now, \begin{align} \left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right) =& \underbrace{\left( y- x\right)^T \nabla f(z)}_{ \text{upper bound using} \ (1) } - \underbrace{\left( y- x\right)^T \nabla f(x)}_{ \text{lower bound using} \ (2) } \\ =& \underbrace{\left( y- x\right)^T \nabla f(z)}_{ \leq f(y) - f(x) + \frac{L}{2} \| x - z \|^2 } - \underbrace{\left( y- x\right)^T \nabla f(x) }_{ \geq f(y) - f(x) - \frac{L}{2} \| x - y \|^2 } \\ \leq & f(y) - f(x) + \frac{L}{2} \| x - z \|^2 - \left[ f(y) - f(x) - \frac{L}{2} \| x - y \|^2 \right] \\ =& \frac{L}{2} \left( \| x - z \|^2 + \| x - y \|^2 \right) \\ \leq& \frac{L}{2} \left( \| x - z \|^2 + \| x - y \|^2 + \underbrace{\| z - y \|^2}_{\geq 0} \right) \end{align}

This completes the proof of $(\clubsuit)$.

Answer 1

What about this simple estimate: $$ \left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right) \le \|y-x\| \cdot \|\nabla f(z)-\nabla f(x)\| \le L \|y-x\| \cdot \|z-x\| \le \frac L2(\|y-x\|^2 + \|z-x\|^2) $$ ??