In the question on Frechet derivative it is used that if $g:\mathbb{R} \to \mathbb{R}$ is a continuous function s.t. $g(x) \in o(x)$, i.e. if $\lim_{x \to 0} \frac{|g(x)|}{x}=0$ then
$$ \int_0^1g(h(t))dt \in o(h) $$
where $h\in C[0,1]$, i.e.
$$ \lim_{||h|| \to 0}\frac{ \int_0^1g(h(t))dt}{||h||} = 0 $$
It is straightforward to check that this holds under the supremum norm in $C[0,1]$: if $\epsilon > 0$ is fixed, then if $\delta>0$ is st. $|g(x)| < \epsilon \cdot |x|$ for all $x \in (-\delta,\delta)$ then for any $h \in C[0,1]$ st. $||h||_{\infty}=\sup_{t\in [0,1]}|h(t)| < \delta$ we have
$$ \Big| \int_0^1g(h(t))dt \Big| \leq \int_0^1|g(h(t))|dt \leq \int_0^1\epsilon|h(t)|dt \leq \epsilon ||h||_{\infty} $$
and thus for any $h\in C[0,1]$ with $||h||_{\infty}<\delta$ we have
$$ \frac{ |\int_0^1g(h(t))dt|}{||h||} \leq \frac{\epsilon ||h||}{||h||} = \epsilon. $$
My question is if it is possible to extend this result to the case where we use the following norm on $C[0,1]$:
$||h|| = \Big(\int |h(t)|^2 dt \Big)^{\frac{1}{2}}$. I have tried to decompose the integral by
\begin{align} \Big| \int_0^1g(h(t))dt \Big| &\leq \Big| \int_{[0,1] \cap (h<\delta) }g(h(t))dt + \int_{[0,1] \cap (h\geq\delta) }g(h(t))dt \Big| \\ &\leq \epsilon\int_0^1h(t)dt + \text{Leb}(\{t\in [0,1] : h(t)\geq \delta\})\cdot C \end{align}
where $C$ is a bound on the continuous function $g \circ h :[0,1] \to \mathbb{R}$. But it does not quite help with proving that the limit is 0.
We don't have any requirements on how $g$ behaves near infinity, and if it can be significantly larger than argument, than it can be possible to find $h$ that is large on segment small enough that $\|h\|$ is still small, but $g \circ h$ is significantly larger than $h$ on this segment.
For example, for $g(x) = x^4$, $\epsilon > 0$ and $$h(x) = \begin{cases} \frac{1}{\epsilon}, x < \epsilon^3\\ \\ \frac{1 - \frac{x - \epsilon^3}{\epsilon^3}}{\epsilon}, \epsilon^3 \leq x < 2 \epsilon^3\\ \\ 0 \end{cases}$$
($1 / \epsilon$ on $[0, \epsilon^3)$, $0$ on $[2\epsilon^3, 1]$ and linear interpolation between)
Then $\|h(x)\| \leq \sqrt{2\epsilon} \to 0$, but $\int_0^1 g(h(x))\, dx > \frac{1}{\epsilon}$.