Exercise 1.1.27 in Tao's Topics in Random Matrices (moments and convergence)

Question

I'm stuck on parts (iii) and (iv). Does anyone have any ideas?

Answer 1

Kavi Rama Murthy gave a good counterexample for $(iv)$ in the comments. Let me give a relatively short argument for $(iii)$.

First, note that by the result of $(ii)$ it is enough to show that $$\limsup_{n \to \infty} \mathbb{E}[|X_n|^k] \leq \mathbb{E}[|X|^k].$$

Now, you can check that for any $K$, $f: x \mapsto |x|^k 1_{\{|x|^k \leq K\}}$ is bounded and upper-semicontinuous and so by the Portmanteau theorem we have that $$\limsup \mathbb{E}[ |X_n|^k 1_{\{|X_n|^k \leq K\}}] \leq \mathbb{E}[|X|^k].$$

Also, by Holder's inequality, you get that $$\mathbb{E}[ |X_n|^k 1_{\{|X_n|^k > K\}}] \leq \mathbb{E}[ |X_n|^{k + \varepsilon}]^{\frac{k}{k+\varepsilon}} \mathbb{P}( |X_n|^k > K)^\frac{\varepsilon}{k + \varepsilon} \leq K^{- \frac{\varepsilon}{k}} \sup_n \mathbb{E}[|X_n|^{k + \varepsilon}] $$ where I have deduced the last inequality by writing $\mathbb{P}(|X_n|^k > K) = \mathbb{P}(|X_n|^{k + \varepsilon} > K^\frac{k + \varepsilon}{k})$ and applying Markov's inequality.

Hence \begin{align*} \limsup \mathbb{E}[|X_n|^k] \leq& \limsup \mathbb{E}[|X_n|^k 1_{\{|X_n|^k \leq K\}}] + \limsup \mathbb{E}[|X_n|^k 1_{\{|X_n|^k > K\}}] \\ \leq & \mathbb{E}[|X|^k] + K^{-\frac{\varepsilon}{k}} \sup_n \mathbb{E}[|X_n|^{k+ \varepsilon}]. \end{align*} It then follows that $\limsup \mathbb{E}[|X_n|^k] \leq \mathbb{E}[|X|^k]$ by sending $K \to \infty$ in the above inequality.