# Does functional satisfies Palais-Smale condition?

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.

• 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 $L^2(0,1)$. – Lorenzo Quarisa Jan 6 at 22:27
• Also note that you made a little mistake when computing the derivative. The correct form is $Df(u_k)[v] = 2 \int_0^{1/2}u_k v dx$. – Danilo Gregorin May 14 at 18:01