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Check if the functional $f(u)=\int_0^{1/2} u^2(x)dx$ satisfies the Palais-Smale condition on the Hilbert space $L^2([0,1],\mathbb{R})$.

We have definied the Palais-Smale condition as follows: $f$ satisfies the PS-condition if every sequence $(u_k)\subset L^2([0,1],\mathbb{R})$ which satisfies $f(u_k)$ is bounded and $Df(u_k) \rightarrow 0$ in $L^2([0,1],\mathbb{R})$, there exists a convergent subsequence.

I think that this functional satisfies the Palais-Smale condition (is this true?) but I don't really know how to prove this correctly. My idea was to take a PS-sequence $(u_k)$, that means $f(u_k)$ is bounded and $Df(u_k) = 2 \int_0^{1/2} u_k(x) \nabla u_k(x) dx \rightarrow 0$ in $L^2([0,1],\mathbb{R})$. And I thought that maybe to prove this, I could show that this PS-sequence is bounded. Then I could follow with Banach-Alaoglu that $u_k$ has a subsequence which converges weakly. Is this idea reasonable or can't I prove the assumption like that?

I wanted to show that $\|u_k\|^2$ is bounded (with the $L^2$-norm) which implies that $u_k$ is bounded and is easier to show, because we know that $\|u_k\|^2 = (u_k,u_k) = \int_0^1 u_k^2(x)dx$. But even if I know that $f(u_k)$ is bounded, I wasn't able to show that the integral is bounded on the whole interval $[0,1]$. Maybe someone could give me some tips to solve this problem.

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    $\begingroup$ I doubt that it satisfies the condition on $L^2(0,1)$. The functional $f$ only detects the behavior of the sequence $\left\{u_k\right\}$ in $(0,1/2)$. So, on $(1/2,1)$ it can do whatever it wants, but to satisfy the $PS$ condition $\left\{u_k\right\}$ needs to have a converging subsequence on $L^2(0,1)$. $\endgroup$ – Lorenzo Quarisa Jan 6 at 22:27
  • $\begingroup$ Also note that you made a little mistake when computing the derivative. The correct form is $Df(u_k)[v] = 2 \int_0^{1/2}u_k v dx$. $\endgroup$ – Danilo Gregorin May 14 at 18:01

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