TITLE: What is the strongest possible statement of the idea that "the tangent line is the best linear approximation"?
QUESTION [13 upvotes]: For instance, I've just checked that that if you take the best linear approximation (in the $L^2$ sense) to a sufficiently nice function $f$ on the interval $[-\varepsilon, \varepsilon]$, and then let $\varepsilon \to 0$, you get $f(0) + x f'(0)$.  
Surely we could make this stronger -- I imagine the analogous statements should hold for, say, the $L^1$ norm as well, or for most reasonable norms.  Can we go farther, though?
Question: What is the strongest precise definition we can give the word "best" so that we have a statement of the form "the tangent line is the best linear approximation to a differentiable function"?  (Feel free to replace "differentiable" with, say, $C^2$ or something if it makes for a more interesting answer.)
(Note: I'm aware of similar-sounding questions here, such as In what sense is the derivative the "best" linear approximation?, but the answers there don't answer my question.)

REPLY [2 votes]: If $f$ is differentiable at $0$ the the same statement holds with uniform approximation instead of $L_2$.
Proof: 
Let $T(x) = f(0) + xf'(0)$.
By the definition of differentiability we have
$$ f(x) - T(x) = \mathcal o(|x|) $$
and thus
$$ \sup_{|x|\le\varepsilon} |f(x) - T(x)| = \mathcal o(\varepsilon).$$
So this would also be satisfied by the best approximation on interval $[-\varepsilon, \varepsilon]$ say $g_\varepsilon$.
In particular, for any mapping $\varepsilon\mapsto x_\varepsilon \in [-\varepsilon, \varepsilon]$ we have
$$ |f(x_\varepsilon) - g_\varepsilon(x_\varepsilon)| \le \sup_{|x|\le \varepsilon} |f(x) - g_\varepsilon(x)| \le \sup_{|x|\le \varepsilon} |f(x) - T(x)| = \mathcal o(\varepsilon).$$
Hence, we have 
$$g_\varepsilon(0) = f(0) + \mathcal o(\varepsilon)$$ 
and
$$ g_\varepsilon' = \frac{g_\varepsilon(\varepsilon) - g_\varepsilon(-\varepsilon)}{2\varepsilon} = \frac{f(\varepsilon) - f(-\varepsilon)}{2\varepsilon} + \mathcal o(1) \to f'(0).$$
For $L_p$ average, $1\le p < \infty$:
Fix some $p\in [1,\infty)$.
For any measurable $\phi$ denote its average $L_p$ norm by 
$$N_\varepsilon \phi = \sqrt[p]{\frac{1}{2\varepsilon} \int_{-\varepsilon}^{\varepsilon} |\phi(x)|^p\,dx}.$$
Let $g_\varepsilon$ be a $L_p$ best approximation on $[-\varepsilon, \varepsilon]$. Then, we also have
$$ N_\varepsilon (f - g_\varepsilon) \le N_\varepsilon (f - T) = \mathcal o(\varepsilon). $$
and
$$ N_\varepsilon (g_\varepsilon - T)
\le N_\varepsilon (f - g_\varepsilon) + N_\varepsilon (f - T) = \mathcal o(\varepsilon). $$