# Sobolev inequality

If $f\in H^2(\mathbb R^2)$, I want to show that

1. $||f||_{L^\infty}\le c||f||_{H^2}$
2. $||f||_{L^\infty}\le c||f||_{H^1} [1+\ln(1+||f||_{H^2})]$

For 1, I use $||f||_{L^\infty}\le \sup_{x\in \mathbb R^2} |f|\le\int_{\mathbb R^2} |\hat f(\xi)|d\xi = \int(1+|\xi|^2)^{-1}(1+|\xi|^2)|\hat f(\xi)|d\xi \le (\int (1+|\xi|^2)^{-2}d\xi)^2(\text {which is integrable in this case})||f||_{H^2}\to \text{is this correct?}$

But for 2, I am stucked. How can I get the "ln"? and how can I make it into a product of $H^1$ and $H^2$ norm?

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To prove 2, try this: First prove that for all $\alpha \in (0,1)$ $$\Vert f \Vert_{L^\infty} \le C_\alpha \Vert f \Vert_{H^1}^\alpha \Vert f \Vert_{H^2}^{1- \alpha}$$ using the Fourier transform and Hölder's inequality. Keep close track of the constant $C_\alpha$. Then minimize the right hand side with respect to $\alpha$. You should get a sharp estimate like $$\Vert f \Vert_{L^\infty}\le c \Vert f \Vert_{H^1} \sqrt{1+\ln(1+ \Vert f \Vert_{H^2})} \:\:\:$$ out of this approach.