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I would refer you to the remark B in supplement 3.4A on Manifolds, Tensor Analysis and Applications of Abraham Marsden Ratiu. It hope it could be useful, and so I quote:

"The following counterexample is due to A.J. Tromba. Let $h: [0, 1] x 1]\times L^2[0, 1]\rightarrow L^2[0, 1]$ be given by $h(x,\phi)=(h'(x))(\phi) =\int_0^1{\sin(\frac{2\pi}{x})\phi(t)} dt$, if $x\neq 0$, and $h(0,\phi)=0$. Continuity at each $x\neq 0$ is obvious and at $x = 0$ it follows by the Riemann-Lebesgue lemma (the Fourier coefficients of a uniformly bounded sequence in $L^2$ relative to an orthonormal set converge to zero). Thus $h$ is $C^0$. However, since $h(x, sin(\frac{2\pi t}{x}))=\frac{1}{2}-\frac{x}{4\pi}\sin(\frac{4\pi}{x})$, we have $h(\frac{1}{n},\sin(2\pi nt))=\frac{1}{2}$ and therefore its $L^2$-norm is $\frac{1}{\sqrt 2}$; this says that $||h'(\frac{1}{n})||\geq \frac{1}{\sqrt 2}$ and thus $h'$ is not continuous."

2 added 4 characters in body

I would refer you to the remark B in section 3.4 supplement 3.4A on Manifolds, Tensor Analysis and Applications of Abraham Marsden Ratiu. It hope it could be useful, and so I quote:

"The following counterexample is due to A.J. Tromba. Let $h: [0, 1] x L^2[0, 1]\rightarrow L^2[0, 1]$ be given by $h(x,\phi)=(h'(x))(\phi) =\int_0^1{\sin(\frac{2\pi}{x})\phi(t)} dt$, if $x\neq 0$, and $h(0,\phi)=0$. Continuity at each $x\neq 0$ is obvious and at $x = 0$ it follows by the Riemann-Lebesgue lemma (the Fourier coefficients of a uniformly bounded sequence in $L^2$ relative to an orthonormal set converge to zero). Thus $h$ is $C^0$. However, since $h(x, sin(\frac{2\pi t}{x}))=\frac{1}{2}-\frac{x}{4\pi}\sin(\frac{4\pi}{x})$, we have $h(\frac{1}{n},\sin(2\pi nt))=\frac{1}{2}$ and therefore its $L^2$-norm is $\frac{1}{\sqrt 2}$; this says that $||h'(\frac{1}{n})||\geq \frac{1}{\sqrt 2}$ and thus $h'$ is not continuous."

1

I would refer you to the remark B in section 3.4 on Manifolds, Tensor Analysis and Applications of Abraham Marsden Ratiu. It hope it could be useful, and so I quote:

"The following counterexample is due to A.J. Tromba. Let $h: [0, 1] x L^2[0, 1]\rightarrow L^2[0, 1]$ be given by $h(x,\phi)=(h'(x))(\phi) =\int_0^1{\sin(\frac{2\pi}{x})\phi(t)} dt$, if $x\neq 0$, and $h(0,\phi)=0$. Continuity at each $x\neq 0$ is obvious and at $x = 0$ it follows by the Riemann-Lebesgue lemma (the Fourier coefficients of a uniformly bounded sequence in $L^2$ relative to an orthonormal set converge to zero). Thus $h$ is $C^0$. However, since $h(x, sin(\frac{2\pi t}{x}))=\frac{1}{2}-\frac{x}{4\pi}\sin(\frac{4\pi}{x})$, we have $h(\frac{1}{n},\sin(2\pi nt))=\frac{1}{2}$ and therefore its $L^2$-norm is $\frac{1}{\sqrt 2}$; this says that $||h'(\frac{1}{n})||\geq \frac{1}{\sqrt 2}$ and thus $h'$ is not continuous."