Sobolev inequality in 1 dimension

Question

Suppose $u \in W^{1,2}(\mathbb{T}^{1})$ with periodic boundary conditions. Then the following Sobolev inequality (in one dimension) holds:

$$ \|u\|_{L^{\infty}} \le C\|u\|_{W^{1,2}}. $$ I have read somewhere that in such a setting, we have by the Sobolev inequality that

$$\|u\|_{L^{\infty}} \le \|u\|_{L^{2}}^{\frac{1}{2}}\|\partial_{x} u\|_{L^{2}}^{\frac{1}{2}}+ \|u\|_{L^{2}},$$ but I cannot see how this follows. The best I can do is \begin{aligned}\|u\|_{L^{\infty}} &\le C\|u\|_{W^{1,2}} = C(\|u\|_{L^{2}}^{2}+\|\partial_{x}u\|_{L^{2}}^{2})^{\frac{1}{2}} \\ &\le C(\|u\|_{L^{2}} + \|\partial_{x}u\|_{L^{2}}) \\ &= C(\|u\|_{L^{2}} + \|\partial_{x}u\|_{L^{2}}^{\frac{1}{2}}\|\partial_{x}u\|_{L^{2}}^{\frac{1}{2}}). \end{aligned}

Are they perhaps using something else other than the inequality I mentioned? I am unable to link the mentioned paper.

Answer 1

Let $u$ be smooth. Then we have $$ u(t)^2 = u(s)^2 +2\int_s^t u(x)u'(x)dx, $$ which implies $$ u(t)^2 \le u(s)^2 + 2\|u\|_{L^2(0,1)}\|u'\|_{L^2(0,1)}. $$ Integrating wrt $s\in (0,1)$ yields $$ u(t)^2 \le \|u\|_{L^2(0,1)}^2 + 2\|u\|_{L^2(0,1)}\|u'\|_{L^2(0,1)}, $$ now take the maximum wrt $t$, to get $$ \|u\|_{L^\infty}^2 \le \|u\|_{L^2(0,1)}^2 + 2\|u\|_{L^2(0,1)}\|u'\|_{L^2(0,1)}. $$ This holds for smooth $u$, and by density, for all $u\in W^{1,2}$.