# Example where $\|\,f_n\|_\infty\to \infty$, but $\|\,f_n\|_1\to 0$

Give an example of a sequence $\,\,f_n : [0, 1] \to \Bbb R$ of continuous functions such that $\,\,\|f_n\|_\infty \to \infty$ but $\int_0^1\lvert\, f_n\rvert\,d\lambda \to 0$ as $n\to\infty$.

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• Think of straight line segments with big $y$-intercept and very small $x$-intercept. – David Mitra Dec 27 '13 at 13:34
• I converted the image to $\TeX$. Please do it yourself next time. Here is a tutorial. – Ayman Hourieh Dec 27 '13 at 13:37

Let $$f_n(x)=\max\left\{n,n^{-1}x^{-1/2}\right\}.$$

Then $\|\,f_n\|_\infty=n$ while $\|\,f\|_1<\dfrac{1}{2n}$.

Here's another: $f_n(x) = nx^{n^2}$

Below I have given a sufficient condition for your problem to have a positive solution in a rather general measure space, afterwhich I give an example in the case of $[0,1]$ with its usual Euclidean topology to help illustrate my construction.

A general solution:

Proposition: More generally, let $X$ be a locally-compact Hausdorff space, $<X,\mathscr{F},\mu>$ be a non-trivial, regular (ie.: $\mu \neq 0$) $\sigma$-finite Borel meaure space which is not purely atomic. Then for any real number $r$, there exists a sequence of measurable functions $f_n$ satisfying: $$f_n\overset{L^{\infty}_m}{\rightarrow} \infty \mbox{ and } f_n\overset{L^{p}_m}{\rightarrow} r ,$$ for every $p\in [1,\infty)$.

Proof: Since the measure space is $\sigma$-finite and non-trivial, there exists some $A\in \mathscr{F}$ with $\mu(A)\leq \infty$. Moreover, since every measure which is not purely atomic is the sum of a purely atomic $\mu_a$ and a purely nonatomic (or diffuse) measure $\mu_d$ (see this article), we may without loss of generality assume $\mu$ to be purely nonatomic (by only considering its nonatomic part).

Now consider $A$ as a disjoint union of two sets $B$ and $C$, both of positive $\mu$-measure.

Now therefore, since $B$ must also be diffuse, for every sufficiently large $k\in \mathbb{Z}^+$ there exists $B_k\in \mathscr{F}$ such that $B_{k+1}\subseteq B_k$ and $\mu(B_k)=\frac{1}{k}. Therefore the sequence of measurable functions: $$g_k:=kI_{A_k} + \frac{r}{\mu(C)}I_{C},$$ must have the desired properties (for large enough$k$). However, the sequence$g_k$is not continous but$C_c(X,\mathbb{R})$is dense in$L_{\mu}^p$, then for every$k \in \mathbb{Z}^+$, there exists some continous function$f_k^p$with compact support such that$\|f_k^p-g_k\|_{L^p_{mu}} = o_{\mu}( 2^{-k})$. Therefore$f_k$converge in$L_{\mu}^p$to r, and in$L^{\infty}_{\mu}$to$\infty$, as desired. A particular case for the Lebesgue measure space Corollary: In the case of the Lebesgue measure space on$[0,2]^d$we may take: $$B_k:=[0,\frac{1}{\sqrt[d]{k}}]^d \mbox{ and } C:=[1,2]^d$$ Then $$f_k(x):= k\prod_{i=1}^d (1-\frac{x-(\frac{1}{\sqrt[d]{k}}+2^{-k})}{2^{-k}})I_{[0,\frac{1}{\sqrt[d]{k}}]}(x_i) + rI_{[1,2]^d}$$. Example: More particularly for$\mathbb{R}$we may take: $$A_k:=[0,\frac{1}{k}]$$ Then$f_k(x):= k(1-\frac{x-(\frac{1}{k}+2^{-k})}{2^{-k}})I_{[0,\frac{1}{k}]}(x_i)solves your problme :). \begin{align*} f_n(x)=\begin{cases}n&\text{ifx\in\left[0,\frac1 {2n^2}\right]$;}\\ \text{[linear descent; see below]}&\text{if$x\in\left(\frac{1}{2n^2},\frac{1}{n^2}\right]$;}\\0&\text{if$x\in\left(\frac1{n^2},1\right].}\end{cases} \end{align*}\|f_n\|_{\infty}=n,\|f_n\|_1<n\times(1/n^2)=1/n$. The linear descent is needed to make the function continuous. It can be computed to be$x\mapsto-2n^3x+2n\$.

[Figure below is not to scale.]