Various convergences of the following sequence

Question

I have the following sequence $(f_n)$ of functions $f_n = \sqrt{n}\,\mathbf{1}_{(0,1/n]}$ where $\mathbf{1}$ is the usual characteristic functions. I have had to evaluate the following three convergence of $(f_n)$, namely

• (1) weak convergence to $0$ in $L^2((0,1))$,
• (2) strong convergence to $0$ in $L^1((0,1))$,
• (3) strong convergence to $0$ in $L^2((0,1))$.

The notions of strong and weak convergence are defined as follows.

Image 1.^src) A sequence $\{f_{n}\}$ of functions in $L^{p}(E)$ for some $1\leq p \leq \infty$, converges in the sense of $L^{p}(E)$ to a function $f \in L^{p}(E)$ if

$$ \lim \| f_{n} - f \|_{p} = 0. $$

This notion of convergence is also called convergence in the mean of order $p$, or in the norm $L^{p}(E)$ or strong convergence in $L^{p}(E)$.

Image 2.^src) Let $1\leq p,q \leq \infty$ be conjugate. A sequence of functions $\{f_{n}\}$ in $L^{p}(E)$ for $1\leq p\leq\infty$, converges weakly to a function $f \in L^{p}(E)$ if

$$ \lim \int_{E} f_n g \, \mathrm{d}\mu=\int_{E}fg \, \mathrm{d}\mu \qquad \text{for all} \quad g \in L^{q}(E). $$

Here are my attempts.

(1) True, since $\lim_{n \to \infty}\int_{(0,1)}\sqrt{n}\,\mathbf{1}_{(0,1/n]}(x) \,dx = \lim_{n \to \infty}\frac{1}{\sqrt{n}}=0$.

(2) True, since $\lim_{n \to \infty} \int_{(0,1)} \left|\sqrt{n}\,\mathbf{1}_{(0,1/n]}(x)\right|\,dx = \lim_{n \to \infty} \int_{(0,1)}\sqrt{n}1_{(0,1/n]}(x)\,dx = 0$.

(3) False, since $\lim_{n \to \infty} \left(\int_{(0,1)} \left|\sqrt{n}\,\mathbf{1}_{(0,1/n]}(x)\right|^2\,dx\right)^{1/2}=1 \neq 0$.

Any help will be greatly appreciated.

Answer 1

For $(1)$ you have to prove that $\int_0^1f_ng \to 0,\forall g \in L^2(0,1)$

It is true,but i do not see(at least now) a quick way to prove this,but in can be proved:

It is true for continuous functions with compact support.

Since this class of functions is dense in $L^p$ spaces,then you can prove it for general $g \in L^2$,using Cauchy-Schwartz,and the denseness of the continuous functions with compact support

I know it is a little bit overkill but it works.

For $(2),(3)$ you are correct.

Answer 2

Supplementing @Marios Gretsas's answer, here is a slightly more direct solution of (1). This is essentially the same as his answer, with a different choice of dense subset of $L^2(0,1)$.

Let $M_n = n^{1/4}$. For any $g \in L^2(0,1)$, write $g = g\mathbf{1}_{\{|g|\leq M_n\}} + g\mathbf{1}_{\{|g|>M_n\}}$. Then

\begin{align*} \left| \int_{0}^{1} f_n(x)g(x) \, \mathrm{d}x \right| &\leq \int_{0}^{1} f_n(x)\left| g(x)\mathbf{1}_{\{|g|\leq M_n\}} \right| \, \mathrm{d}x + \left| \int_{0}^{1} f_n(x)g(x)\mathbf{1}_{\{|g|>M_n\}} \, \mathrm{d}x \right| \\ &\leq \frac{M_n}{\sqrt{n}} + \left\|g \mathbf{1}_{\{|g|>M_n\}} \right\|_{2}, \end{align*}

where we utilized the Cauchy-Schwarz inequality together with $\|f\|_{2}=1$ in the last line. By invoking the Dominated Convergence Theorem, we check taht $\left\|g \mathbf{1}_{\{|g|>M_n\}} \right\|_{2} \to 0$. So this bound converges to $0$ as and hence $\int_{0}^{1} f_n(x)g(x) \, \mathrm{d}x \to 0$.