Let $F(x)$ be a continuous (not necessarily monotonic) function defined on smooth curve $C$. I am wondering if the following inequality holds for line integrals.
$$|F(a)-F(b)|\leq \int_C ||f(x)||dx\leq \sup_{x\in C}|| f(x)||\cdot L_C$$
where $f(x)=\nabla F(x)$ is the gradient of $F(x)$, $L_C$ is the length of the curve $C$ and $||\cdot||$ is some norm (e.g., $\ell_2$ norm). We also assume that $\displaystyle \sup_{x\in C}|| f(x)||$ is bounded.
Thanks.
Consider $F : {\bf R}^n\rightarrow {\bf R}$ and any curve $c:[0,1]\rightarrow {\bf R}^n,\ t\mapsto (x_1(t),\cdots, x_n(t))$ where $c(0)=p,\ c(1)=q$. Then consider a function $ g(t):= (F\circ c)(t)$ Here $\circ$ means composite : $ (F\circ c)(t) = F(c(t))$. That is $g : [0,1] \rightarrow {\bf R}$. So recall the fundamental theorem in calculus. We will apply to $g$ : $$ g(1)-g(0)= \int_0^1 g'(t) dt\ \ast $$
Here $g$ is composite function. So we compute at details. First by change rule, $$ g'(t) =\sum_{k=1}^n \frac{\partial }{\partial x^k} F \frac{d}{dt} x_k (t) $$
Usually we denote it : $$ g'(t) = dF(c'(t)) = \nabla F\cdot c'(t) $$ In the first $dF$ is $1\times n$-matrix and in the second, the gradient $\nabla F$ is a vector. In fact matrix multiplication is equal to inner product in this case.
And $$ F(q)-F(p)= g(1)-g(0)= \int_0^1 g'(t)dt = \int_0^1 \nabla F\cdot c'(t) dt $$ so that $$| F(q)-F(p)| \leq \int_0^1 |\nabla F| |c'(t)| dt \leq {\rm sup} |\nabla F| \cdot {\rm length} (c)$$
Here we used $|\int k(t) dt|\leq \int|k(t)|dt$, $ |v\cdot w|\leq |v||w|$ and $\int_0^1 |c'(t)|dt = {\rm length} (c)$