step in proof for gagliardo-nirenberg interpolation

119 Views Asked by At

I am having difficulty with the proof of proposition 12.84 of A First Course in Sobolev Spaces, 2nd Edition, by Giovanni Leoni. In particular, the following intermediate result is proven: $$|u(x)| \le (\alpha_N R^N)^{-1/q}\|u\|_{L^q} + R^\alpha|u|_{C^{o,\alpha}}\tag{1}$$ where $u$ is Holder Continuous with constants $|u|_{C^{0,\alpha}}$ and $\alpha > 0$, $R>0$, $\|\cdot\|_{L^q}$ is the $L^q$ norm, $\alpha_N$ is a constant, and $x\in \mathbb{R}^N$.

From here, the proof then minimizes the right hand side with respect to $R$. In other words, they minimize the following function: $$g(t) = \alpha_N^{-1/q}t^{-N/q}\|u\|_{L^q} + t^\alpha|u|_{C^{o,\alpha}}\tag{2}$$ and get the following by inserting the minimum of (2) into (1). $$\|u\|_{L^\infty} \le c\|u\|^{\theta_1}_{L^q}|u|^{1-\theta_1}_{C^{0,\alpha}} \tag{3}$$ where $\theta_1 = \alpha/(\alpha + N/q)$ and $c$ is a grouping of constant terms dependent on $\alpha, N,$ and $q$. Since (1) is taken over all $x\in\mathbb{R}^R$, I see where the $\|u\|_{L^\infty}$ term comes from. However, I am having difficulty seeing how the right hand side is derived.

In particular, when I take the derivative of (2) and set it equal to 0, I get the following: $$0 = g'(t) = -\frac{N}{q} \alpha_N^{-1/q} t^{-N/q - 1}\|u\|_{L^q} + \alpha t^{\alpha - 1}|u|_{C^{0,\alpha}} \tag{6}$$ which after a couple of algebra steps yields

$$t = \left(\frac{\alpha q |u|_{C^{0,\alpha}}}{\alpha_N^{-1/q}N\|u\|_{L^q}}\right)^{\alpha+N/q} = a\left(\frac{|u|_{C^{0,\alpha}}}{ \|u\|_{L^q}}\right)^{\alpha + N/q}\tag{4}.$$

Plugging $t$ in (4) into $R$ in (1) yields $$(\alpha_N t^N)^{-1/q}\|u\|_{L^q} + t^\alpha|u|_{C^{o,\alpha}} = \alpha_N^{-1/q}a|u|^{\alpha+N/q}\|u\|^{1-\alpha-N/q} + a|u|^{\alpha+N/q+1}\|u\|^{-\alpha-N/q}. \tag{5} $$ From (5), I am not sure how to get to get to (3).

Edit:

To go from (6) to (4), the following steps are taken: $$\frac{N}{q}\alpha_N^{-1/q}\|u\|_{L^q}t^{-N/q-1} = \alpha t^{\alpha - 1}|u|_{C^{0,\alpha}}$$ $$\frac{N}{q}\alpha_N^{-1/q}\|u\|_{L^q}t^{-N/q-\alpha} = \alpha |u|_{C^{0,\alpha}}$$ $$t^{-N/q-\alpha} = \frac{\alpha q |u|_{C^{0,\alpha}}}{\alpha_N^{-1/q}N \|u\|_{L^q}}$$ $$t = \left(\frac{\alpha q |u|_{C^{0,\alpha}}}{\alpha_N^{-1/q}N \|u\|_{L^q}}\right)^{\alpha + N/q}$$

1

There are 1 best solutions below

0
On BEST ANSWER

Your last step contains an error. It should read $$ t = (\ldots)^{-1/(\alpha + N/q)}.$$