S.-T. Yau College Student Mathematics Contests 2019 This problem is from S.-T. Yau College Student Mathematics Contests 2019.
Prove that there exists a universal constant $K$,   for  all $C^1$ function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, if  $f\in L^{1}\left(
\mathbb{R}^{2}\right) \cap L^2\left( \mathbb{R}^{2}\right) $ and $|\nabla f| \in L^2(\mathbb{R}^2)$, we have the following inequality:
$$
\left\Vert f\right\Vert _{L^{2}\left( \mathbb{R}^{2}\right) }^{2}\leq
K\left\Vert f\right\Vert _{L^{1}\left( \mathbb{R}^{2}\right) }\left\Vert
\nabla f\right\Vert _{L^{2}\left( \mathbb{R}^{2}\right) }\text{.}
$$
Can you provide constant $K$ so that $K<10$? In the problem, all the $L^p$-spaces are defined with respect to the Lebesgue measure.
 A: The result is trivial if either $\Vert f \Vert_{L^1(\mathbb{R}^2)} = 0$ or $\Vert \nabla f \Vert_{L^2(\mathbb{R}^2)} = 0$, as in either case the hypotheses require that $f =0$.  We may assume, then, that both of these quantities are positive.
We use Plancherel to compute
$$
\int_{\mathbb{R}^2} |f(x)|^2 dx =  \int_{\mathbb{R}^2} |\hat{f}(\xi)|^2 d\xi.
$$
Then for any $r >0$ we split the integral and bound
$$
\int_{\mathbb{R}^2} |\hat{f}(\xi)|^2 d\xi = \int_{B(0,r)} |\hat{f}(\xi)|^2 d\xi + \int_{B(0,r)^c} |\hat{f}(\xi)|^2 d\xi \\
\le \Vert \hat{f} \Vert_{L^\infty(\mathbb{R}^2)}^2 \cdot \pi r^2 + \frac{1}{4\pi^2 r^2} \int_{B(0,r)^c} 4\pi^2 |\xi|^2 |\hat{f}(\xi)|^2 d\xi \\
= \pi r^2 \Vert \hat{f} \Vert_{L^\infty(\mathbb{R}^2)}^2  + \frac{1}{4\pi^2 r^2} \int_{\mathbb{R}^2} |\widehat{\nabla f}(\xi)|^2 d\xi \\
= \pi r^2 \Vert \hat{f} \Vert_{L^\infty(\mathbb{R}^2)}^2  + \frac{1}{4\pi^2 r^2}\Vert \nabla f\Vert_{L^2(\mathbb{R}^2)}^2.
$$
In the last two steps we've used the usual diagonalization of differentiation by the Fourier transform and another application of Plancherel.
Now, for $A,B >0$ consider the function $g: (0,\infty) \to (0,\infty)$ defined by $g(r) = Ar^2 + B/r^2$.  It's easy to show this has a global minimum at $r$ such that
$$
0 = g'(r) = 2 A r -2 \frac{B}{r^3} \Leftrightarrow r = \frac{B^{1/4}}{A^{1/4}},
$$
and so the minimum value is
$$
g_{min} = g(\frac{B^{1/4}}{A^{1/4}}) = 2A^{1/2} B^{1/2}. 
$$
We apply this lemma to our above bound with $A = \pi \Vert \hat{f}\Vert_{L^\infty(\mathbb{R}^2)}^2$ and $B = \Vert \nabla f \Vert_{L^2(\mathbb{R}^2)}^2 / (4 \pi^2)$ to see that
$$
\int_{\mathbb{R}^2} |f(x)|^2 dx \le 2 \left(\pi \Vert \hat{f}\Vert_{L^\infty(\mathbb{R}^2)}^2 \right)^{1/2} \left(\frac{\Vert \nabla f \Vert_{L^2(\mathbb{R}^2)}^2 }{ 4 \pi^2} \right)^{1/2} = \frac{1}{\sqrt{\pi}} \Vert \hat{f}\Vert_{L^\infty(\mathbb{R}^2)} \Vert \nabla f \Vert_{L^2(\mathbb{R}^2)}.
$$
Finally, we use the bound $\Vert \hat{f}\Vert_{L^\infty(\mathbb{R}^2)} \le \Vert f\Vert_{L^1(\mathbb{R}^2)}$ to arrive at the desired estimate:
$$
\Vert f \Vert_{L^2(\mathbb{R}^2)}^2 \le \frac{1}{\sqrt{\pi}} \Vert f\Vert_{L^1(\mathbb{R}^2)} \Vert \nabla f \Vert_{L^2(\mathbb{R}^2)}.
$$
A: Using Fourier transform, it is sufficient to prove that
$$\|f\|_{L^2}^2\leq K \|f\|_{L^{\infty}} \||x|f\|_{L^{2}}.$$
Let $A_{n}:= B_{2^{n+1}} (0) \setminus B_{2^{n}}(0) $ denote the annulus, and divide
$f=\Sigma f_{n}$ with supp $f_{n}\in A_{n},$
note that
$$\int f_{n}^2 \leq \|f\|_{L^{\infty}} \int f_{n} \leq C2^n \|f\|_{L^{\infty}}\|f_{n}\|_{L^{2}}.$$
Combine with the fact $\||x|f_n\|_{L^{2}}\geq 2^n\|f_n\|_{L^{2}},$ we have
$$\|f_{n}\|_{L^2}^2\leq C \|f\|_{L^{\infty}} \||x|f_{n}\|_{L^{2}}.$$
On the other hand
$$\|f\|_{L^2}^2=\Sigma_{n} \|f_{n}\|_{L^2}^2 \leq  C \|f\|_{L^{\infty}} \Sigma_{n}\||x|f_{n}\|_{L^{2}}.$$
Set $a_{n}=\||x|f_{n}\|_{L^{2}},$ then $\||x|f\|_{L^{2}}^2=\Sigma_{n} a_{n}^2,$
the basic inequality gives the result. You can choose the best constants to prove that $K<10$. Anyway, 快乐就完事了.
