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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!

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How is $Hf(0)$ defined? – copper.hat Dec 12 '12 at 7:13
I'm sorry if this happens to be irrelevant to this particular question (I haven't done it myself). But the Ascoli-Arzela theorem is very useful when you're proving compactness of operators on $C([0,1])$. – Gyu Eun Lee Dec 12 '12 at 7:16
@copper.hat: That is not said in my exercise. Will it matter? anonymous. : Thanks! But will it help me now when I will prove not compactness and problably compactness in $L^2$? – Johan Dec 12 '12 at 7:38
Possibly. I've got no clue, I'm only really just getting into functional analysis myself. – Gyu Eun Lee Dec 12 '12 at 7:49
@copper.hat: By continuity, $Hf(0)$ has to be defined as $f(0)$. – 23rd Dec 12 '12 at 8:40
up vote 7 down vote accepted

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.

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Great, why if $(F_n)$ equicontinuous? – Johan Dec 12 '12 at 9:44
@Johan: It is because $|f_n|\le 1$ implies $|F_n(x)-F_n(y)|\le|x-y|$. – 23rd Dec 12 '12 at 9:46
Thanks and why can we take the n limit before the $\delta$ – Johan Dec 12 '12 at 9:58
@Johan: $\delta$ and $n$ are independent of each other. If you are not familiar with this trick, you may use the definition of limit and $\epsilon-\delta/\epsilon-N$ language to fill in the details. – 23rd Dec 12 '12 at 10:01
That seems reasonable, you know anywhere there is a proof of that? – Johan Dec 12 '12 at 10:22

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