Prove that $\|f\|_p \leq \liminf \|f_n\|_p$ under weak convergence

Question

Let $1<p<\infty$ and $q$ its conjugate. Given a sequence $(f_k)_{k \in \mathbb N}$ and $f$ in $L^p(\mathbb R^d)$, I am trying to show that if for all $g \in L^q(\mathbb R^d)$, $$\lim_{k \to \infty} \int_{\mathbb R^d}f_k(x)g(x)dx=\int_{\mathbb R^d} f(x)g(x)dx,\tag1$$ then $$||f||_p \leq \lim \inf ||f_k||_p$$

I couldn't prove this, I've tried to use the property $||h||_p=\sup_{||g||_q \leq 1}|\int_{\mathbb R^d}h(x)g(x)dx|$

By (1), if we take supremum over $g$ such that $||g||_q \leq 1$, we have $$||f||_p=\sup_{||g||_q \leq 1} \lim_{k \to \infty} \int_{\mathbb R^d}f_k(x)g(x)dx$$

I don't know how to relate this to the limit inferior.

Answer 1

Assume $p \ge 1$. (What is the conjugate of $p$ if $0 < p < 1$?)

If $g \in L^q$ and $\|g\|_q \le 1$ then $$\left| \int f_k g \, dx \right| \le \|f_k\|_p$$ by Holder's inequality. Thus $$\liminf_{k \to \infty} \left| \int f_k g \, dx \right| \le \liminf_{k \to \infty} \|f_k\|_p.$$ But $$\liminf_{k \to \infty} \left| \int f_k g \, dx \right| = \lim_{k \to \infty} \left| \int f_k g \, dx \right| = \left| \int fg \, dx \right|$$ so that $$\left| \int fg \, dx \right| \le \liminf_{k \to \infty} \|f_k\|_p.$$ Now take the supremum over all such $g$ to get $$\|f\|_p \le \liminf_{k \to \infty} \|f_k\|_p.$$

Answer 2

If you have a Banach space $X$ then $x_n\to x$ in the weak topology by definition means that $f(x_n)\to f(x)$ for every $f\in X^{\ast}$. By Hahn-Banach's theorem $||x_n||=\sup_{f\in X^{\ast};|f|=1}|f(x_n)|\geq |f(x_n)|$. Taking the $\liminf$ you have

$$\liminf_n ||x_n||\geq f(x).$$

Again by Hahn-Banach's theorem and taking $\sup$ on $|f|=1$ you have that

$$\liminf ||x_n||\geq ||x||.$$

In your case $\infty>p\geq 1$ and so for $X=L^p$ you have $X^{\ast}=L^q$.