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Which assumption together with compactness imply this? Weak convergence in X or strong convergence in Y or both? Could you elaborate more?

Let X be compactly embedded in Y. Assume also that there is a sequence f_n in X such that f_n converges to f weakly in X and strongly in Y to some function f in X. Can we say that f_n converges to f strongly in X?

I would say the solution is some kind of a line on the xy=1 plane in 3D. Your solution would be a closed form of an equation. So I don't know if saying that {(x,1/x,z) satisfying the equation x^4+1/x^4-z^4-4z^2-2=0 } is the solution set is enough.

Because it will be an approximation not really the solution itself.

Your mistake is in the multiplication (-f + kx2 +BDx2) * ( -k -BD ). On the right hand side rather than -fBD you should get +fBD

which numerical method are you using?

It looks correct.

None of your links work.

I didn't get what you mean. Are you asking it as a question?

By Schur's lemma every weakly Cauchy sequence converges. So your answer lies in the proof of Schur's lemma.

I seem to suggest using series all the time but the Taylor series expansion of e^h might be useful. Then (e^h-1)/h = 1+h(some terms) tends to 1 as h goes to 0. But this may again be circular.

Maybe you could expand e^{4x} in its Taylor series expansion. and then look at the integral of (e^{4x}-1)/x+1/x. In the first term there will be a cancellation of a power of x so it will be a polynomial integration. and the second gives ln(x). But depending on your integration bound there might...

Hi i have a question about L^2 spaces and convergence. Here it goes: Let K\subset \mathbb{R}^2 be bounded. Let g,h\in L^2(K), and a sequence f_n\in L^2(K) such that f_n converges strongly to f\in L^2. Is it true that \lim_{n\rightarrow \infty} \int_{K} f_n g h = \int_{K} f g h? If it is...