Best linear approximation in a neighborhood property for Frechet derivative Let $D$ denote a Frechet derivative at a point $a$. Then for $f:\mathbb{R}\to\mathbb{R}$ it holds for all $x$ from some neighborhood of $a$:
$$|f(x)-(f(a)+D(x-a))|\leq|f(x)-(f(a)+M(x-a))|,\ M\neq D$$
Circular definition of tangent line and derivative
Prove that $T(x)$ is the most accurate linear approximation to $f(x)$ near $x_0$...
Do we have this for Frechet derivative in other normed spaces $f:\mathbb{S}\to\mathbb{P}$ also?
I can only get that for some $\delta$ and all $x$ s.t. $||x-a||_S\leq \delta$
$$||f(x)-(f(a)+D(x-a))||_{P}\leq\frac{1}{3}||D-M||||x-a||_S$$
$$\operatorname{sup}_{||x-a||_S\leq\delta}||f(x)-(f(a)+D(x-a))||_{P}\leq\frac{1}{3}\delta||D-M||$$
$$\delta||D-M||=\operatorname{sup}_{||x-a||_S\leq\delta}||(D-M)(x-a)||_P\leq\frac{1}{3}\delta||D-M||+$$
$$+\operatorname{sup}_{||x-a||_S\leq\delta}||f(x)-(f(a)+M(x-a))||_{P}$$
So it follows
$$\operatorname{sup}_{||x-a||_S\leq\delta}||f(x)-(f(a)+D(x-a))||_{P}<\operatorname{sup}_{||x-a||_S\leq\delta}||f(x)-(f(a)+M(x-a))||_{P}$$
But this is a weaker result of course.
Upd: changed the result to strict inequality.
 A: Ok, it seems that in general the best local linear approximation doesn't need to exist.
I found the counterexample here If there is a linear function $g$ which is at least as good of an approximation as any other linear $h$, then $f$ is differentiable at $x_0$. in Milo Brandt's answer and comments.
Let $f(x,y)=x^2+y^2$. Then the linear approximation with the derivative at point $(0, 0)$ is $l(x,y)=0$. But the function $g(x,y)=x$ coincides with $f(x,y)$ at points along the curve $x=\frac{1-\sqrt{1-4y^2}}{2}$ in any neighborhood of $(0,0)$ because of continuity. So the linear approximation with the derivative is not the best linear approximation in any neigborhood of $(0,0)$.
Another note.
Existence of the local best linear approximation in the sense of suprema defined in the question doesn't imply differentiability. Taking $f(x)=|x|$ it can be seen that for $g(x)=0$ for any $\delta>0$ and for all $|x|<\delta$:
$$\operatorname{sup}_{|x|<\delta}|f(x)-g(x)|=\operatorname{sup}_{|x|<\delta}|x|=\delta$$
And for any $l(x)=mx$ with $m\neq0$ it holds
$$\operatorname{sup}_{|x|<\delta}|f(x)-l(x)|=\operatorname{max}\{\delta|1+m|;\delta|1-m|\}>\delta=\operatorname{sup}_{|x|<\delta}|f(x)-g(x)|$$
But $|x|$ is not differentible at $0$.
