Sobolev estimates $\|\phi\|_2\leq\|\phi\|_{H^2}$

Question

I have been reading the paper 'Convergence of a finite volume scheme for a system of interacting species with cross-diffusion' by Carrillo et al. in Section 4.1 they have been using an estimate,

$\|\nabla\phi\|_{L^{\infty}(\Omega)}\leq \|\phi\|_{H^2(\Omega)}$,

which seems quite intuitive, however I am unable to show the result. I already know for a bounded domain, $\Omega$

$\|\cdot\|_{L^1(\Omega)}\leq\|\cdot\|_{L^2(\Omega)}$,

and this can be extended to any finite $p$. However, this is not true for the essential supremum norm. $\|\cdot\|_{L^p(\Omega)}\nleq\|\cdot\|_{L^{\infty}(\Omega)}$, as pointed out by @falcon

I also unsuccessfully tried to use the Poincaré Inequality, stated as $\|\phi\|_{L^p}\leq\|\nabla\phi\|_{L^p}$, for $\phi$ in $H^2(\Omega)$.

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EDIT: This assertion is false as shown by the counterexamples provided by @falcon. I am posting a new question with the correct form as given by @calvin-khor

Answer 1

Well, this is not true. Take $\Omega = [0,1/2]$ and $\phi(x) = 2$ for all $x$, then $$\|\phi\|_{L^\infty} = 2$$ but $$\|\phi\|_{H^2} = \left(\|\phi\|_{L^2}^2 + \|\phi'\|_{L^2}^2 + \|\phi''\|_{L^2}^2\right)^{1/2} = \int_0^{1/2}2 = 1. $$

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EDIT: Still not true. Take $\Omega = [0,1/2]$ and $\phi(x) = 2x$, then $$\|\nabla \phi\|_{L^\infty} = 2$$ but \begin{align} \|\phi\|_{H^2} &= \left(\|\phi\|_{L^2}^2 + \|\phi'\|_{L^2}^2 + \|\phi''\|_{L^2}^2\right)^{1/2} \\ &= \left(4\int_0^{1/2} x^2 dx + 4\int_0^{1/2} dx\right)^{1/2}\\ &= \left(\frac{1}{6}+ 2\right)^{1/2} < 2. \end{align}