Operators on $C([0,1])$ that is compact or not. For $f\in C([0,1])$ set
$$Hf(x) = \frac{1}{x}\int_0^x f(t)dt.$$
a) Show that $H$ is a bounded operator from $C([0,1])$ into itself which is not compact.
b) From a) it follows that $H$ induces a bounded linear map from $C([0,1])$ to $L^2([0,1])$. Is this map compact?  
My try: a) $|H| \leq 1$ so it must be continuous and into  $C([0,1])$. For the not compactness I need to find a sequence $(Hf_n(x))$ with no cauchy subsequences? 
Any hints would be great!
 A: a) Let 
$$f_n(x)=\left\{\begin{array}{cl} 2^nx, & x\in[0,2^{-n}]\\1\ ,& x\in[1-2^{-n},1]\end{array}\right..$$ Then $f_n\in C([0,1])$ and $\|f_n\|_{C([0,1])}=1$. Denote $g_n=Hf_n$. Then 
$$g_n(x)=\left\{\begin{array}{cl} 2^{n-1}x, & x\in[0,2^{-n}]\\1-2^{-n-1}x^{-1},& x\in[1-2^{-n},1]\end{array}\right..$$
Note that if $m<n$, then $g_m(2^{-n})=2^{m-n-1}\le\frac{1}{4}$, but $g_m(2^{-m})=\frac{1}{2}$, which implies that $(g_n)$ has no Cauchy subsequence.
b) $H:C([0,1])\to L^2([0,1])$ is compact. Let $(f_n)$ be a sequence in the closed unit ball of $C([0,1])$ and denote $g_n=Hf_n$. It suffices to show that  $(g_n)$ has convergent subsequence in $L^2([0,1])$.
Denote $F_n(x)=\int_0^xf_n(t)dt$. Since $(f_n)$ is uniformly bounded, $(F_n)$ is uniformly bounded and equicontinous. Then by Arzelà–Ascoli theorem, $F_n$ has some subsequence convergent in $C([0,1])$. Without loss of generality, let us assume that $\lim_{n\to\infty}F_n=F$ in $C([0,1])$, and we only need to show that $g_n$ converges to $g(x)=F(x)/x$ in $L^2([0,1])$.
Note that $|F_n(x)|\le x$ for every $x\in[0,1]$, so $|F(x)|\le x$, $|g_n(x)|\le 1$ and $|g(x)|\le 1$. Then for every $\delta>0$,
$$\int_0^1|g_n(x)-g(x)|^2dx= \int_0^\delta|g_n(x)-g(x)|^2dx+\int_\delta^1|g_n(x)-g(x)|^2dx$$ 
$$\le 4\delta+\delta^{-2}\int_\delta^1|F_n(x)-F(x)|^2dx.$$
First letting $n\to\infty$ and then letting $\delta\to 0$, the conclusion follows.
