The directional derivative of function $ F(f) $ is by definition the old-fashioned
$$
D_uF(f) = \lim_{\epsilon\to 0} {{F(f+u\epsilon)-F(f)}\over{\epsilon}}
$$
$$
= \lim_{\epsilon\to 0} {{\langle\langle f+u\epsilon, g\rangle\rangle - \langle\langle f,g \rangle \rangle}\over{\epsilon}}
$$
Recall that the inner product of two functions $ \langle\langle f,g \rangle\rangle = \int_0^1{f(x) \cdot g(x)dx}$ , we have
$$
D_uF(f) = \lim_{\epsilon\to 0} {{\int_0^1{(f+u \epsilon)(x)\cdot g(x)dx} - \int_0^1{f(x) \cdot g(x)dx}}\over{\epsilon}}
$$
$$
= \lim_{\epsilon\to 0} {{\int_0^1{\epsilon u(x)\cdot g(x)dx} }\over{\epsilon}}
$$
$$
= {\int_0^1{u(x) \cdot g(x)dx}}
$$
$$ = \langle\langle u, g\rangle\rangle
$$
Since $ \langle\langle\nabla F(f), u\rangle\rangle = D_uF(f) $ must be true for all u, we get
$$
\nabla F(f) = g
$$
keenan
@taoy1 Terrific. One way you could make this writeup a little clearer is by adding just a couple more steps for your limit. I.e., show how this expression ends up becoming just $\langle\langle u, g \rangle\rangle$.
The directional derivative of function $ F(f) $ is by definition the old-fashioned $$ D_uF(f) = \lim_{\epsilon\to 0} {{F(f+u\epsilon)-F(f)}\over{\epsilon}} $$ $$ = \lim_{\epsilon\to 0} {{\langle\langle f+u\epsilon, g\rangle\rangle - \langle\langle f,g \rangle \rangle}\over{\epsilon}} $$ Recall that the inner product of two functions $ \langle\langle f,g \rangle\rangle = \int_0^1{f(x) \cdot g(x)dx}$ , we have $$ D_uF(f) = \lim_{\epsilon\to 0} {{\int_0^1{(f+u \epsilon)(x)\cdot g(x)dx} - \int_0^1{f(x) \cdot g(x)dx}}\over{\epsilon}} $$ $$ = \lim_{\epsilon\to 0} {{\int_0^1{\epsilon u(x)\cdot g(x)dx} }\over{\epsilon}} $$ $$ = {\int_0^1{u(x) \cdot g(x)dx}} $$ $$ = \langle\langle u, g\rangle\rangle $$ Since $ \langle\langle\nabla F(f), u\rangle\rangle = D_uF(f) $ must be true for all u, we get $$ \nabla F(f) = g $$
@taoy1 Terrific. One way you could make this writeup a little clearer is by adding just a couple more steps for your limit. I.e., show how this expression ends up becoming just $\langle\langle u, g \rangle\rangle$.
Thanks. I have edited my writeup above.