Is the gradient of $\langle z, F(\cdot)\rangle$ Lipschitz continuous when the one of $F$ is?

This is my first topic here, please apologise if I am doing something wrong.

Let $X$ and $Y$ be Banach spaces with their duals $X^*$ and $Y^*$, respectively, and $F:X\to Y$ with the Frechet gradient Lipschitz continuous with constant $L$. Let $z\in Y^*$ with $\|z\|=1$ (can be also $\leq 1$ or equal to some given constant). What can be said about the Lipschitz continuity of the gradient of the function defined by $x \mapsto \langle z , F(x) \rangle$? One has $∇ \langle z, F(x) \rangle = z \circ ∇F(x)$, but from here I am lost and I need to show that $∇ \langle z, F(x) \rangle$ is Lipschitz continuous and to determine the corresponding Lipschitz constant.

Thank you all in advance for any hint or answer, which will be credited in my Master' Thesis.

• – Shaun Aug 29 '17 at 20:19
• I tried editing but I couldn't figure out the last two lines – tattwamasi amrutam Aug 29 '17 at 20:21
• $\nabla G(x_1) - \nabla G(x_2) = z \circ \bigl(\nabla F(x_1) - \nabla F(x_2)\bigr)$ – Daniel Fischer Aug 29 '17 at 20:27
• thanks, I'll ty this – Tobias Wu Aug 29 '17 at 21:14
• @DanielFischer: you mean that one can write ||∇G(x)−∇G(y)||=||z∘(∇F(x)−∇F(y))|| $\leq$ ||z|| ||∇F(x)−∇F(y)|| $\leq$ L||x-y||, so the answer were that the function is Lipschitz continuous with L as the constant? – Tobias Wu Aug 29 '17 at 22:10