Directional derivatives of $f \mapsto \max f$ Consider the functional $\Psi \colon C^{0}([0,1]) \to \mathbb R$ defined by 
$$
\Psi(f):=\max_{x \in [0,1]} f(x)
$$
Find the directional derivative (if it exists) in the generic point $f$ in the generic direction $g$.
I should find the limit
$$
\frac{\partial \Psi}{\partial g}(f):=\lim_{t \to 0} \frac{\max(f+tg)-\max f}{t}
$$
but I do not know how to estimate $\max(f+tg)$. Thanks.  
 A: Let $K = \{ x \in [0, 1] : f(x) = \max f \}$ be the set of maximum points. Then $K$ is a non-zero compact subset of $[0, 1]$. We want to prove that
$$ D_{g}^{+}\Psi(f) := \lim_{t \to 0^{+}} \frac{\Psi(f+tg) - \Psi(f)}{t} = \max_{K} g \tag{1} $$
and
$$ D_{g}^{-}\Psi(f) := \lim_{t \to 0^{-}} \frac{\Psi(f+tg) - \Psi(f)}{t} = \min_{K} g. \tag {2}$$
Assuming (1) and (2), we know that $D_{g}\Psi(f)$ exists if and only if $\max_{K}g = \min_{K}g$, or equivalently, $g$ is constant on the set $K$. Moreover, if (1) is proved, then (2) follows straightforward from the relation
$$D_{g}^{-}\Psi(f) = - D_{-g}^{+}\Psi(f). $$
So it suffices to prove (1).
Proof. We prove that both the limsup and the liminf coincide. Estimation for liminf is easy part. Choose $x \in K$ so that $\max_{K}g = g(x)$. Then $f(x) + tg(x) = \Psi(f) + t\max_{K}g$ and we have
$$ \frac{\Psi(f + tg) - \Psi(f)}{t} \geq \max_{K}g
\quad \Longrightarrow \quad
\liminf_{t\to 0^{+}} \frac{\Psi(f + tg) - \Psi(f)}{t} \geq \max_{K}g. $$
So let us estimate the limsup. To this end, let $(t_{n})$ be any sequence satisfying $t_{n} \to 0^{+}$, and for each $n$ we choose $x_{n}$ such that $f(x_{n}) + t_{n}g(x_{n}) = \Psi(f + t_{n}g)$. Since $\Psi (f + t_{n}g) \geq \Psi(f) - t_{n}\| g\|_{\infty}$, we get
$$ x_{n} \in \{ x \in [0, 1] : f(x) \geq \Psi(f) - t_{n}\| g\|_{\infty} \} =: K_{n}. $$
Here, note that $K = \bigcap_{n=1}^{\infty} K_{n}$. But since
$$
\frac{\Psi(f + t_{n}g) - \Psi(f)}{t_{n}}
= \frac{f(x_{n}) + t_{n}g(x_{n}) - \Psi(f)}{t_{n}}
\leq g(x_{n})
\leq \max_{K_{n}} g,$$
it follows that
$$ \limsup_{n\to\infty} \frac{\Psi(f + t_{n}g) - \Psi(f)}{t_{n}}
\leq \limsup_{n\to\infty} g(x_{n})
\leq \lim_{n\to\infty} \max_{K_{n}} g
= \max_{K}g.$$
Since this is true for any sequence $t_{n} \to 0^{+}$, it follows that
$$ \limsup_{t \to 0^{+}} \frac{\Psi(f + tg) - \Psi(f)}{t} \leq \max_{K} g. $$
This completes the proof.
