The embedding $H^2(\mathbb{R})$ into $L^4(\mathbb{R})\cap L^6(\mathbb{R})$

Question

In one paper, the auther says the sobolev space $H^2(\mathbb{R})$ can be embedded into $L^4(\mathbb{R})\cap L^6(\mathbb{R})$. However, by checking the sobolev embedding theorem, I couldn't find any corresponding to this.

Answer 1

Specifically in your 1-dimensional situation: If $f\in H^2(\mathbb{R})$, then for any $x\in \mathbb{R}$ we have $$ \int_x^{x+1} |f(t)|\,dt \le \sqrt{\int_x^{x+1} |f(t)|^2\,dt} \le \|f\|_{H^2} $$ hence there exists $y\in [x,x+1]$ such that $|f(y)|\le \|f\|_{H^2}$. By the Fundamental Theorem of Calculus, $$ |f(x)-f(y)| \le \int_x^y |f'(t)|\,dt \le \sqrt{\int_x^y |f'(t)|^2\,dt } \le \|f\|_{H^2} $$ In conclusion, $\|f\|_{L^\infty(\mathbb{R})}\le 2\|f\|_{H^2}$.

And since $f\in L^2(\mathbb{R})$, the standard interpolation of Lebesgue spaces shows $f\in L^p(\mathbb{R})$ for $2\le p\le \infty$, all with linear bounds $\|f\|_{L^p}\le C_p \|f\|_{H^2}$.

Generally

When the Sobolev exponent is above the space dimension $(p>n)$, we get $W^{1,p}(\mathbb{R}^n)$ embedded in $L^\infty(\mathbb{R}^n)$. Indeed, for any $x_0$ consider the unit ball $B$ centered at $x_0$ and apply Morrey's inequality there, getting a bound on $\sup_B |f|$.