Example where $\|\,f_n\|_\infty\to \infty$, but $\|\,f_n\|_1\to 0$ Please help me to solve the following problem that is in the Lebesgue integral discussion 

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 $n\to\infty$.

Original screenshot
 A: Let 
$$
f_n(x)=\max\left\{n,n^{-1}x^{-1/2}\right\}.
$$
Then $\|\,f_n\|_\infty=n$ while $\|\,f\|_1<\dfrac{1}{2n}$.
A: 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:
\begin{equation}
f_n\overset{L^{\infty}_m}{\rightarrow} \infty 
\mbox{ and }
f_n\overset{L^{p}_m}{\rightarrow} r ,
\end{equation}
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:
\begin{equation}
g_k:=kI_{A_k} + \frac{r}{\mu(C)}I_{C},
\end{equation} 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: \begin{equation}
B_k:=[0,\frac{1}{\sqrt[d]{k}}]^d \mbox{ and } C:=[1,2]^d
\end{equation}
Then 
\begin{equation}
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}
\end{equation}.  
Example: More particularly for $\mathbb{R}$ we may take: \begin{equation}
A_k:=[0,\frac{1}{k}]
\end{equation}
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 :).  
A: Here's another: $f_n(x) = nx^{n^2}$
A: \begin{align*}
f_n(x)=\begin{cases}n&\text{if $x\in\left[0,\frac1 {2n^2}\right]$;}\\
\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.]

