It is false that "Since $[-1,1]$ is compact, the $L^{1}$-norm
is equivalent to $L^{\infty}$-norm."
The following is a famous counter-example (in probability theory). (one may argue that the given $f_n$ are not continuous, but we can always smooth out the function at the edge.)
Let $f_{1}=1_{[0,1]}$, $f_{2}=1_{[0,\frac{1}{2}]}$, $f_{3}=1_{[\frac{1}{2},1]}$,
$f_{4}=1_{[0,\frac{1}{4}]}$, $f_{5}=1_{[\frac{1}{4},\frac{2}{4}]}$,
$f_{6}=1_{[\frac{3}{4},\frac{4}{4}]},$$f_{7}=1_{[0,\frac{1}{8}]}$,
$\ldots$.
Clearly $||f_{n}||_{1}\rightarrow0$ but $||f_{n}||_{\infty}=1$ for
all $n$.
The correct way to solve the problem:
Firstly, we show that $(f_{n})$ is a Cauchy sequence with respect
to the $L^{1}$-norm. Let $m,n\in\mathbb{N}$ with $m<n$. By direct
calculation, we have that
$$
|f_{n}(x)-f_{m}(x)|\leq2\cdot1_{[0,\frac{1}{m}]}(x).
$$
That is, $|f_{n}(x)-f_{m}(x)|$ vanishes if $x\in[-1,0]\cup[\frac{1}{m},1]$
and $|f_{n}(x)-f_{m}(x)|\leq2$ if $x\in[0,\frac{1}{m}]$. Therefore,
$\int_{-1}^{1}|f_{n}-f_{m}|\,d\lambda\leq\frac{2}{m}$. Hence, $(f_{n})$
is a Cauchy sequence with respect to the $||\cdot||_{1}$-norm.
Next, we show that $(f_{n})$ does not converges to any element in
$C[-1,1]$. Prove by contradiction. Suppose there exists $f\in C[-1,1]$
such that $\int_{-1}^{1}|f_{n}-f|\,d\lambda\rightarrow0$. By Vitali
Theorem, there exists a subsequence $(f_{n_{k}})$ and a measurable
set $A\subseteq[-1,1]$ with $\lambda([-1,1]\setminus A)=0$ such
that for each $x\in A$, $|f_{n_{k}}(x)-f(x)|\rightarrow0$ as $k\rightarrow\infty$.
(i.e., $f_{n_{k}}-f\rightarrow0$ pointwisely a.e.).
For $x\in A$, clearly, $\lim_{k}f_{n_{k}}(x)=1_{(0,1]}(x)$. Hence,
$f(x)=1_{(0,1]}(x)$ for $x\in A$. Since $f$ is continuous and $\lambda([-1,1]\setminus A)=0$,
it follows that $f(x)=0$ for $x\in[-1,1]\setminus A$ too. (For,
$[-1,1]\setminus A$ has measure zero $\Rightarrow$ it cannot contain
any interior point $\Rightarrow$ for each $x\in[-1,1]\setminus A$,
there exists a sequence $(x_{n})$ in $A$ such that $x_{n}\rightarrow x$.
By the continuity of $f$...).
That is, $f=1_{(0,1]}$, which is a contradiction.
Best Answer
Let $\varepsilon > 0$ be fixed. Note that \begin{eqnarray*} \int_{0}^{1/2} \left| (x + 1/2)^n - (x + 1/2)^m \right| \, dx & \leq & \int_{0}^{1/2} \left| (x + 1/2)^n \right| \, dx + \int_{0}^{1/2} \left| (x + 1/2)^n \right| \, dx \\ & = & \left[\frac{1}{n+1} - \frac{(1/2)^{n+1}}{n + 1} \right] + \left[ \frac{1}{n+1} - \frac{(1/2)^{m+1}}{m + 1} \right]\\ & < & \frac{1}{n} + \frac{1}{m}. \end{eqnarray*} From here, can you determine what constraints you must put on $n$ (resp. $m$) so tha $\dfrac{1}{n} < \dfrac{\varepsilon}{2}$?