Let $1<p<\infty$, $f_{k}\in L^{p}(\mathbb{S}^{1})$, $f\in L^{p}(\mathbb{S}^{1})$ be such that $f_{k}\longrightarrow f$ weakly in $L^{p}(\mathbb{S}^{1})$. That is for any $g\in L^{p'}(\mathbb{S}^{1})$, where $p'=\frac{p}{p-1}$, we have $$\int_{-\pi}^{\pi}f_{k}gdx\longrightarrow\int_{-\pi}^{\pi}fgdx\ \text{as}\ k\longrightarrow\infty.$$
Then, the exercise state three parts
(a) If $p=2$, show that $|f_{k}|^{2}-|f|^{2}-|f_{k}-f|^{2}\longrightarrow 0\ \text{in the sense of distribution}.$
(b) If $p\neq 2$, do we still have $|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p}\longrightarrow 0\ \text{in the sense of distribution}?$ Prove it or give a counterexample.
(c) Assume $p\neq 2$, $f_{k}\longrightarrow f$ weakly in $L^{p}(\mathbb{S}^{1})$ and $f_{k}\longrightarrow f$ a.e., can you show $|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p}\longrightarrow 0\ \text{in the sense of distribution}?$
(I deleted my previous writings, since the attempt is misleading and I will answer my own post for part (a) and part (b). This will also make this post more clear).
Edit 1: Bounty
I have no problem with $(a)$ and $(b)$ now. The solution given by "StarBug" for $(b)$ is not correct. The example he provided converges in $Lp$ too nicely, and he did not explain why does it imply $|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p}$ does not converge to $0$ in the sense of distribution.
I came up another one and I will post it by answering my own question in a few days. Also, for $(a)$, we need to assume they are complex, so we need to take complex conjugate, but the proof is still similar.
The bounty is for (c). I still have no idea for (c). One hint I got from my professor today is that part (c) is an important result for Calculus of Variation, but I don't really know anything about this...
Edit 2:
Okay. I figured them all out. I just posted the proof by answering my own question.
Again, the answer that got most upvotes is wrong.
Okay, I figured it out about 1 week ago, below is the proof for all three parts of this exercise.
Proof of part (a):
The proof can be a little bit easier if we assume all the things are real, but let us by more general and consider the case in which they are complex.
Firstly using $|g|=g\overline{g}$, we can simplify $$|f_{k}|^{2}-|f|^{2}-|f_{k}-f|^{2}=f_{k}\overline{f_{k}}-f\overline{f}-(f_{k}-f)\overline{(f_{k}-f)}=f_{k}\overline{f_{k}}-f\overline{f}-(f_{k}-f)(\overline{f_{k}}-\overline{f})=-2|f|^{2}+f_{k}\overline{f}+\overline{f_{k}}f,$$ so that for any test function $\phi$, we have $$\int_{-\pi}^{\pi}(|f_{k}|^{2}-|f|^{2}-|f_{k}-f|^{2})\phi dx=\int_{-\pi}^{\pi}f_{k}(\overline{f}\phi)dx+\int_{-\pi}^{\pi}\overline{f_{k}}(f\phi)dx-\int_{-\pi}^{\pi}2|f|^{2}\phi dx\ \ \ (*).$$
Note that $p=2\implies p'=2$.
To estimate the first term above, note that since $f\in L^{2}(\mathbb{S}^{1})$, and $|f|=\Big|\overline{f}\Big|$, we must have $\int_{-\pi}^{\pi}\Big|\overline{f}\Big|^{2}dx=\int_{-\pi}^{\pi}|f|^{2}dx<\infty,$ and thus $\overline{f}\in L^{2}(\mathbb{S}^{1})$ as well. Also, $C^{\infty}\subset L^{p}$, so $\phi\in L^{2}(\mathbb{S}^{1})$. Thus, it follows from Cauchy Schwarz that $\overline{f}\phi\in L^{2}(\mathbb{S}^{1})=L^{p'}(\mathbb{S}^{1})$. Therefore, $$\int_{-\pi}^{\pi}f_{k}(\overline{f}\phi)dx\longrightarrow \int_{-\pi}^{\pi}f(\overline{f}\phi)dx=\int_{-\pi}^{\pi}|f|^{2}\phi dx\ \text{as}\ k\longrightarrow\infty.$$
To estimate the second term, with the similar argument, we know that $\overline{f_{k}}\in L^{2}(\mathbb{S}^{1})=L^{p}(\mathbb{S}^{1})$ and $f\phi\in L^{2}(\mathbb{S}^{1})=L^{p'}(\mathbb{S}^{1})$. Also, since $f_{k}$ converges weakly to $f$ in $L^{p}$, it must be that $$\lim_{k\rightarrow\infty}\int_{\pi}^{\pi} \overline{f_{k}}\overline{g}dx=\overline{\lim_{k\rightarrow\infty}\int_{-\pi}^{\pi} f_{k}gdx}=\overline{\int_{-\pi}^{\pi}fgdx}=\int_{-\pi}^{\pi}\overline{f}\overline{g}dx,$$ but by the similar argument we know that $g\in L^{p'}\implies \overline{g}\in L^{p'}$, and thus $\overline{f_{k}}\longrightarrow \overline{f}$ weakly in $L^{p}$.
Hence, $$\int_{-\pi}^{\pi}\overline{f_{k}}(f\phi)dx\longrightarrow \int_{-\pi}^{\pi}\overline{f}(f\phi)dx=\int_{-\pi}^{\pi}|f|^{2}\phi dx\ \text{as}\ k\longrightarrow\infty.$$
Using these two estimates in $(*)$ yields us the desired result: $$\int_{-\pi}^{\pi}(|f_{k}|^{2}-|f|^{2}-|f_{k}-f|^{2})\phi dx\longrightarrow\int_{-\pi}^{\pi}|f|^{2}\phi dx+\int_{-\pi}^{\pi}|f|^{2}\phi dx-\int_{-\pi}^{\pi}2|f|^{2}\phi dx=0\ \text{as}\ k\longrightarrow\infty,$$ whence $|f_{k}|^{2}-|f|^{2}-|f_{k}-f|^{2}\longrightarrow 0$ as $k\longrightarrow\infty$, in the sense of distribution.
Proof of part $(b)$:
The answer is No. We need a small lemma for us to construct the counterexample.
Proof of Lemma:
Firstly, it is clear that $\sin(kx)\in L^{p}(\mathbb{S}^{1})$ for any $1< p<\infty.$ The rest of the proof is then a standard application of Riemann-Lebesgue Lemma.
Indeed, fix a $1<p<\infty$, and set $p':=p/(p-1)$. Note that since $\mathbb{S}^{1}$ has finite lebesgue measure, it is a standard fact from measure theory that $L^{q}(\mathbb{S}^{1})\subset L^{1}(\mathbb{S}^{1})$ for any $1< q<\infty.$ In particular, since $0<p-1< p\implies p'=p/(p-1)>p/p=1,$ we have $L^{p'}(\mathbb{S}^{1})\subset L^{1}(\mathbb{S}^{1})$.
Let $g\in L^{p'}(\mathbb{S}^{1})$, then by above argument, $g\in L^{1}(\mathbb{S}^{1})$. It then follows from Riemann-Lebesgue Lemma that $$\int_{-\pi}^{\pi}\sin(kx)g(x)dx=\dfrac{1}{2 i}\int_{-\pi}^{\pi}g(x)e^{ikx}dx-\dfrac{1}{2 i}\int_{-\pi}^{\pi}g(x)e^{-ikx}dx\longrightarrow 0,\text{as}\ k\longrightarrow\infty.$$
Therefore, $\sin(kx)$ converges weakly to $0$ in $L^{p}(\mathbb{S}^{1})$ for all $1<p<\infty$, as desired.
Consider functions defined by $f_{k}(x):=1+\sin kx$ and $f(x):=1$. By the above lemma, we know that $f_{k}(x)\longrightarrow f$ weakly in $L^{p}$ as $k\longrightarrow\infty$ for any $p>1$.
Consider the case of $p=4$ and the test function $\phi_{*}:=1$. With a little bit of computation, we have $$|f_{k}|^{4}-|f|^{4}-|f_{k}-f|^{4}=4\sin^{3}(kx)+6\sin^{2}(kx)+4\sin(kx),$$ so that \begin{align*} \int_{\pi}^{\pi}(|f_{k}|^{4}-|f|^{4}-|f_{k}-f|^{4})\phi_{*} dx&=\int_{\pi}^{\pi}(|f_{k}|^{4}-|f|^{4}-|f_{k}-f|^{4})dx\\ &=\int_{-\pi}^{\pi}4\sin^{3}(kx)dx+\int_{-\pi}^{\pi}6\sin^{2}(kx)dx+\int_{-\pi}^{\pi}4\sin(kx)dx\\ \end{align*}
For the first and the third terms, we know that $\sin^{3}(kx)$ and $\sin(kx)$ are both odd functions and $[\pi,\pi]$ is symmetric about $0$, and thus $$\int_{-\pi}^{\pi}4\sin^{3}(kx)dx=0=\int_{-\pi}^{\pi}4\sin(kx)dx.$$
For the second one, using the trick $\sin^{2}(u)=\frac{1}{2}-\frac{1}{2}\cos(2u)$ and then doing the standard calculus yield that $$\int_{-\pi}^{\pi}6\sin^{2}kxdx=6\pi-\dfrac{3\sin(2\pi k)}{k}\longrightarrow 6\pi\ \text{as}\ k\longrightarrow\infty.$$ The limiting behavior is a direct result from a standard application of the Squeeze Theorem.
Therefore, for $\phi_{*}=1$, we see that $$\int_{\pi}^{\pi}(|f_{k}|^{4}-|f|^{4}-|f_{k}-f|^{4})\phi_{*} dx\longrightarrow 6\pi\ \text{as}\ k\longrightarrow \infty,$$ which implies that $\int_{\pi}^{\pi}(|f_{k}|^{4}-|f|^{4}-|f_{k}-f|^{4})\phi dx$ does NOT converge to $0$ for ALL test functions $\phi$, and thus $|f_{k}|^{4}-|f|^{4}-|f_{k}-f|^{4}$ does NOT converge to $0$ in the sense of distribution.
Proof of $(c)$:
Let $0<\epsilon<1$, and define $A_{k,\epsilon}:=\{x\in [-\pi,\pi]:|f(x)|<\epsilon|f_{k}(x)|\}.$ Let $\phi$ be a test function, we then split the integral as follows $$\Big|\int_{-\pi}^{\pi}(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\phi dx\Big|\leq \Big|\int_{A_{k,\epsilon}}(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\phi dx\Big|+\Big|\int_{A_{k,\epsilon}^{c}}(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\phi dx\Big|.$$
Firstly, we need below three preliminaries results to evaluate the first integral:
$(1)$ Since $f_{k}$ converges to $f$ weakly in $L^{p}$, it follows immediately from Uniform Boundedness Principle that there exists a finite constant $C>0$ such that $\|f_{k}\|_{L^{p}}\leq C$ for all $k$;
$(2)$ On $A_{k,\epsilon}$, we have $|f|/|f_{k}|<\epsilon\implies -\epsilon<f/f_{k}<\epsilon;$
$(3)$ The function $1-x^{p}-(1-x)^{p}$ is nondecreasing and positive for all $0<x<\epsilon$ and $\epsilon$ sufficiently small. On the other hand, $1-(-x)^{p}-(1-x)^{p}$ is negative but still non-decreasing for all $-\epsilon<x<0$.
Using the above three results, we are able to evaluate as follows: \begin{align*} \Big|\int_{A_{k,\epsilon}}(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\phi dx\Big|&\leq \int_{A_{k,\epsilon}}\Big|(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\Big|\cdot |\phi| dx\\ &\leq \|\phi\|_{\infty}\int_{A_{k,\epsilon}}\Big|(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\Big|dx\\ &=\|\phi\|_{\infty}\int_{A_{k,\epsilon}}|f_{k}|^{p}\Big|\Big(1-\dfrac{|f|^{p}}{|f_{k}|^{p}}-\Big|1-\dfrac{f}{f_{k}}\Big|^{p}\Big)\Big|dx\\ &\leq \|\phi\|_{\infty}\int_{A_{k,\epsilon}}|f_{k}|^{p}\max\{1-\epsilon^{p}-(1-\epsilon)^{p}, (1+\epsilon)^{p}-1+\epsilon^{p}\}dx\\ &=\|\phi\|_{\infty}\max\{1-\epsilon^{p}-(1-\epsilon)^{p}, (1+\epsilon)^{p}-1+\epsilon^{p}\}\int_{A_{k,\epsilon}}|f_{k}|^{p}dx\\ \end{align*} \begin{align*} \Big|\int_{A_{k,\epsilon}}(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\phi dx\Big|&=\|\phi\|_{\infty}\max\{1-\epsilon^{p}-(1-\epsilon)^{p}, (1+\epsilon)^{p}-1+\epsilon^{p}\}\int_{A_{k,\epsilon}}|f_{k}|^{p}dx\\ &\leq \|\phi\|_{\infty}\max\{1-\epsilon^{p}-(1-\epsilon)^{p}, (1+\epsilon)^{p}-1+\epsilon^{p}\}\int_{-\infty}^{\infty}|f_{k}|^{p}dx\\ &=\|\phi\|_{\infty}\|f_{k}\|_{L^{p}}\max\{1-\epsilon^{p}-(1-\epsilon)^{p}, (1+\epsilon)^{p}-1+\epsilon^{p}\}\\ &\leq C\|\phi\|_{\infty}\max\{1-\epsilon^{p}-(1-\epsilon)^{p}, (1+\epsilon)^{p}-1+\epsilon^{p}\}\longrightarrow 0\ \text{as}\ \epsilon\searrow 0. \end{align*}
Several details in the above equations are now explained below:
$(1)$ In the $4^{th}$ line, since by $(2)$ we know that $-\epsilon<f/f_{k}<\epsilon$, applying $(3)$ can create the inequality:
$(1a)$ $1-x^{p}-(1-x)^{p}$ is nondecreasing for $0<x<\epsilon$, so we have $\leq (1-\epsilon^{p}-(1-\epsilon)^{p});$
$(1b)$ $1-(-x)^{p}-(1-x)^{p}$ is nondecreasing for $-\epsilon<x<0$, so we have $$1-(+\epsilon)^{p}-(1+\epsilon)^{p}\leq 1-(-x)^{p}-(1-x)^{p}\leq 0.$$
$(2)$ In the $7^{th}$ line, I expanded the integral, and the expansion is consistent with $\leq$ since the integrand is non-negative (is in the absolute value);
$(3)$ In the last line, I used $(1)$.
To evaluate (the absolute value of) the second integral, we want to apply the dominated convergence theorem to $(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\mathbb{1}_{A_{k,\epsilon}^{c}}.$ To do so, firstly note that since $f_{k}\longrightarrow f$ almost surely, it is then clear that almost surely we have $(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\mathbb{1}_{A_{k,\epsilon}^{c}}\longrightarrow 0\ \text{as}\ k\rightarrow\infty.$
Also on $A_{k,\epsilon}^{c}$ we have $|f_{k}|/|f|\leq\epsilon$, and thus \begin{align*} (|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\mathbb{1}_{A_{k,\epsilon}^{c}}\phi&\leq \|\phi\|_{\infty}\Big|(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\Big|\\ &\leq\|\phi\|_{\infty}\Big(|f_{k}|^{p}+|f|^{p}+2^{p-1}|f_{k}|^{p}+2^{p-1}|f|^{p}\Big)\\ &=\|\phi\|_{\infty}|f|^{p}\Big(\dfrac{|f_{k}|^{p}}{|f|^{p}}+1+2^{p-1}\dfrac{|f_{k}|^{p}}{|f|^{p}}+2^{p-1}\Big)\\ &\leq \|\phi\|_{\infty}|f|^{p}(\epsilon+1+2^{p-1}+\epsilon 2^{p-1})\\ &=C_{\epsilon, p}\|\phi\|_{\infty}|f|^{p}\in L^{1}. \end{align*}
Hence, it follows from that \begin{align*} \Big|\int_{A_{k,\epsilon}^{c}}(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\phi dx\Big|&\leq \|\phi\|_{\infty}\int_{-\pi}^{\pi}\Big|(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\mathbb{1}_{A_{k,\epsilon}^{c}}\Big|dx\longrightarrow 0\ \text{as}\ k\longrightarrow\infty. \end{align*}
This two evaluates together gives us $$\Big|\int_{-\pi}^{\pi}(|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p})\phi dx\Big|\longrightarrow 0\ \text{as}\ k\longrightarrow\infty.$$
Since this holds for arbitrary test function, we are able to conclude that $|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p}\longrightarrow 0$ as $k\longrightarrow\infty$ in the sense of distribution.
The proof of this exercise is then concluded.