$||f_{n}||_{0}=\sup|xf_{n}(x)|$, $f_{n}\in{C[0,1]}$.

Question

Suppose $f_{n}\in{C[0,1]}$

$||f_{n}||_{0}=\sup|xf_{n}(x)|$, where $x\in{[0,1]}$.

$||f_{n}||_{1}=\int_{0}^{1} |f_{n}(x)| dx$.

I want to find an example, such that,

$||f_{n}||_{0}\rightarrow{0}$ and $||f_{n}||_{1}\nrightarrow{0}$

Anyone has some ideas?

Answer 1

Hint: as an element of $C(0,1]$, we have $\|\frac1{nx}\|_0=\frac1n$ and $\|\frac1{nx}\|_1=\infty$. With some minor manipulation (change what the function looks like on $[0,\epsilon)$ for some small $\epsilon>0$) you can make it into an example that would work for you.

Answer 2

Consider the following sequence $$ f_n(x) = \frac{c_n}{x+1/n}\in C^0([0,1]), $$ Where $c_n = \frac{1}{\int_0^1 \frac{1}{x+1/n}dx}$. Then, $\int_0^1 f_n(x)=1$. We see that $$ x+1/n \geq x+1/(n+1) \quad \iff \quad \frac{1}{x+1/n} \leq \frac{1}{x+1/(n+1)}, $$ whic implies that $\frac{1}{x+1/n}$ is an increasing sequence that converges pointwisely almost everywhere to $\frac{1}{x}$. By monotone convergence theorem (Beppo-Levi) $$ \lim_n \int_0^1 \frac{1}{x+1/n}dx = \int_0^1 \lim_n \frac{1}{x+1/n}dx = \int_0^1 \frac{1}{x}dx = \infty. $$ This shows that $c_n \to 0$ as $n\to\infty$. But then $$ \sup_{x\in[0,1]} |xf_n(x)| = c_n \sup_{x\in[0,1]} \frac{x}{x+1/n} \leq c_n \cdot 1 = c_n \to 0. $$