Show that $\|\check{f_k}\|_1 \to \infty$ as $k \to \infty$ Define $$f_k(x)=\chi_{[-1,1]}*\chi_{[-k,k]}$$ 
(a) Compute $f_k(x)$ explicitly and show that $\|f\|_{u}=2$
(b) Show that $\check{f_k}=(\pi x)^{-2}\sin 2\pi kx\sin2\pi x$ and $\|\check{f_k}\|_1 \to \infty$ as $k \to \infty$
For (a)
  $$f_k(x)=\int\chi_{[-1,1]}(y) \chi_{[-k,k]}(x-y) dy=\int\chi_{[-1,1]\cap[-k+x,k+x]}(y) dy=m\left([-1,1]\cap[-k+x,k+x]\right)$$
Thus, $$ f_k(x) = \left\{ \begin{array}{ll}
         0 & \mbox{if $x \leq -1-k$};\\
        x+k+1 & \mbox{if $-1-k\le x \le 1-k$}\\
2& \mbox{if $1-k\le x \le k-1$}\\
1+k-x&\mbox{$k-1 \le x \le 1+k$}\\
0&\mbox{$x \ge 1+k$}.\end{array} \right.  $$
It is clear from the above is $\|f\|_u=2$. 
For part (b), 
We know that $f_k(x)=\chi_{[-1,1]} * \chi_{[-k,k]}$ and ${f_k}(x)=\hat{f_k}(-x)$. Now $\hat{f_k}(x)=\hat{\chi_{[-1,1]}}\hat{\chi_{[-k,k]}}$
Now $$\hat{\chi_{[-k,k]}}(x)=\int\chi_{[-k,k]}(y)e^{-2i\pi xy}dy=\int_k^ke^{-2i\pi xy}dy=\frac{\sin 2\pi kx}{\pi x}$$
This gives that $$\hat{f_k}(x)=(\pi x)^{-2}\sin 2\pi kx\sin2\pi x$$ and hence $\check{f_k}=(\pi x)^{-2}\sin 2\pi kx\sin2\pi x$. 
Now $$\|\check{f_k}\|_1=\int \frac{1}{(\pi x)^2} |\sin 2\pi x||\sin 2k\pi x|dx$$ which after substituting $y=2k\pi x$ becomes $$\frac{2k}{\pi}\int\frac{1}{y^2}|\sin(\frac{y}{k})||\sin y| dy$$
I don't know how to show that this goes to $\infty$ as $k \to \infty$. 
Any hint will be greatly appreciated. Thanks in advance!!
 A: Edit: the previous computation was wrong, see Alonso's comment below.
Consider the integral before the substitution, and estimate $$\begin{align*}\int_{\mathbb R}\frac{1}{\pi^2x^2}|\sin(2\pi x)||\sin(2k\pi x)|\,dx&\geq \int_0^{1/4}\frac{1}{\pi^2 x^2}|\sin(2\pi x)||\sin(2k\pi x)|\,dx\\
&\geq C\int_0^{1/4}\frac{2\pi x}{\pi^2x^2}|\sin(2k\pi x)|\,dx\\
&=C\int_0^{1/4}\frac{1}{x}|\sin(2k\pi x)|\,dx.\end{align*}$$ We now set $2kx=y$, and estimate $$\begin{align*}\int_0^1\frac{1}{x}|\sin(2k\pi x)|\,dx&=\int_0^{k/2}\frac{1}{y}|\sin(\pi y)|\,dy\\
&\geq\sum_{j=0}^{[k/2]-1}\int_{2j+1/4}^{2j+3/4}\frac{1}{y}|\sin(\pi y-2j\pi)|\,dy.\end{align*}$$ But, when $y\in\left[2j+1/4,2j+3/4\right]$, then $$\pi y-2j\pi\in\left[\frac{\pi}{4},\frac{3\pi}{4}\right]\Rightarrow \sin(\pi y-2j\pi)\geq\frac{\sqrt{2}}{2},$$ therefore $$\begin{align*}\int_0^1\frac{1}{x}|\sin(2k\pi x)|\,dx&\geq\sum_{j=0}^{[k/2]-1}\int_{2j+1/4}^{2j+3/4}\frac{1}{y}|\sin(\pi y-2j\pi)|\,dy\\
&\geq\frac{\sqrt{2}}{2}\sum_{j=0}^{[k/2]-1}\int_{2j+1/4}^{2j+3/4}\frac{1}{y}\,dy\\
&\geq\frac{\sqrt{2}}{2}\sum_{j=0}^{[k/2]-1}\int_{2j+1/4}^{2j+3/4}\frac{1}{2j+3/4}\,dy\\
&=\frac{\sqrt{2}}{4}\sum_{j=0}^{[k/2]-1}\frac{1}{2j+3/4}.\end{align*}$$ As $k\to\infty$, the last series behaves like the harmonic series, therefore the $L^1$ norms go to infinity as $k\to\infty$.
