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The picture below is taken from this paper: http://real.mtak.hu/22877/.

The authors claim that the basis of $H^2(\Omega) \cap H^1(\Omega)$ denoted by $\lbrace w_i \rbrace _{i \geq 1}$ can be extended to be a basis of $L^2(\Omega;H^1(0,1))$. I don't see how it can be possible. In my thinking, we have to multiply $w_i(x)$ by $h_i(x)$ where $h_i(x)=\frac{cos(i \pi x )}{i\pi}$ is a basis of $H^1(0,1)$. Is this right?. Thank you.

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    $\begingroup$ where is the contradiction? You can do what you said and obtain what they said. They just imposed the value at $\rho=0$ (calling $\rho$ the second variable). Your statement stands, up to a scaling by $i\pi$ (it cannot be $x$ and $x$) : $ w_i(x)h_i(\rho)i \pi$ works fine for example. $\endgroup$
    – username
    Commented May 2, 2021 at 17:54
  • $\begingroup$ Thank you sir for the answer. $\endgroup$
    – Gustave
    Commented May 2, 2021 at 18:01

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