Spaces of class $J_\alpha$

Question

This question is about the spaces of class $J_\alpha$.
Given three Banach spaces $Z\subset Y\subset X$ (with continuous embeddings), and given $\alpha\in (0,1)$, we say that $Y$ is of class $J_\alpha$ between $X$ and $Z$ if there is $C>0$ such that \begin{equation} \left\|y\right\|_Y\leq C \left\|y\right\|_Z^\alpha\left\|y\right\|_X^{1-\alpha}, \quad y\in Z. \end{equation} My question is: Can we say that the space $C[0,1]$ is of class $J_\alpha$ between $H^2(0,1)$ and $L^2(0,1)$. It is obvious that $H^2(0,1)\subset C[0,1] \subset L^2(0,1)$. In addition, one can show that there is $\tilde{C}>0$ such that \begin{equation} \left\|y\right\|_{\infty}\leq \tilde{C} \left\|y\right\|_{H^2}, \end{equation} which in a way reflects the embedding $H^2(0,l) \subset C[0,l]$. Also, we trivially have \begin{equation} \left\|y\right\|_{L^2}\leq \left\|y\right\|_{\infty}. \end{equation}

I am not sure if the original assumption is correct and would be really grateful for any suggestions!

Answer 1

One approach is to integrate by parts. This works nicely when $y(0)=y(1)=0$, because then $$\int_0^1 (y')^2 =-\int_0^1 y\,y''\le \|y\|_{L^2}\|y''\|_{L^2}\le \|y\|_{L^2}\|y\|_{H^2} \tag1$$ which implies $$\|y\|_{L^\infty} \le \|y'\|_{L^1} \le \|y'\|_{L^2} \le \|y\|_{L^2}^{1/2}\|y\|_{H^2}^{1/2} \tag2$$ In general one has to deal with boundary terms, which is possible but less pleasant.

"you should totally drop that and try jQuery Fourier series"

Indeed, let's write $y(t)=\sum c_n e^{2\pi int}$; here and below summation is over ${n\in\mathbb Z}$. The norms are roughly $\|y\|_{L^2}=(\sum |c_n|^2)^{1/2}$ and $\|y\|_{H^2}=(\sum (|n|+1)^4 |c_n|^2)^{1/2}$. Clearly, $\|y\|_\infty\le \sum |c_n|$. The problem reduces to the numerology of exponents in Hölder's inequality (a.k.a. Hölderology). For example,
$$\sum |c_n| = \sum |c_n|^{1/2}|c_n|^{1/2} \le \left(\sum |c_n|^2\right)^{1/4} \left(\sum |c_n|^{2/3}\right)^{3/4} \tag3$$ and then $$\sum |c_n|^{2/3} = \sum (|n|+1)^{-4/3} (|n|+1)^{4/3}|c_n|^{2/3} \\ \\ \le \left(\sum (|n|+1)^{-2}\right)^{2/3} \left( \sum (|n|+1)^{4}|c_n|^{2} \right)^{1/3} \tag4$$ The end result is $$\begin{split}\sum |c_n| &\le \left(\sum |c_n|^2\right)^{1/4} \left(\sum (|n|+1)^{-2}\right)^{1/2} \left( \sum (|n|+1)^{4}|c_n|^{2} \right)^{1/4} \\ &\le C\|y\|_{L^2}^{1/2}\|y\|_{H^2}^{1/2} \end{split} \tag5$$

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There is some slack in (5): it uses $\sum(|n|+1)^{-2}<\infty$, but $\sum (|n|+1)^{-r}$ converges for any $r>1$. You may want to redo the Hölderology to find $\alpha<1/2$ for which $\sum |c_n| \le C\|y\|_{L^2}^{1-\alpha}\|y\|_{H^2}^{\alpha}$ holds. Can you find the smallest possible $\alpha$?

Here is a lower bound for $\alpha$, by means of the example $y(t)=((\epsilon-t)^+)^2$. Here $\|y\|_\infty = \epsilon^2$, $\|y\|_{L^2}\approx \epsilon^{5/2}$ and $\|y\|_{H^2}\approx \epsilon^{1/2}$. Therefore, $$\epsilon^2 \le C \epsilon^{\alpha/2}\epsilon^{5(1-\alpha)/2}$$ for arbitrarily small $\epsilon$, which implies $\alpha\ge 1/4$.