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I have to show that the following family of functions: \begin{align*} \mathfrak{F}= \left\{f \in C\left([-1,1]\right) : \int_{-1}^{1}f(x)dx \in [0,1]\right\} \end{align*}

is closed; but that it is not bounded nor equicontinuous.

I think I've managed to show that it is closed. (proof at the end, revisions are welcome!) But I'm really struggling to find a counterexample for the bounded part! How can a function defined on a bounded domain with a bounded integral, not be bounded?

Could anyone please share some examples where this happens!

Proof that the family is closed:

Suppose $f \in \overline{\mathfrak{F}}$, then there exists $\left(f_{k_n}\right)_{n=1}^{\infty} \subseteq \mathfrak{F}$ such that: \begin{align*} ||f_{k_n} - f||_{[-1,1]} \rightarrow 0 \quad \mathrm{when:} n\ \rightarrow 0 \end{align*}

then as $f_{n_k} \rightarrow f$ uniformly, we can then assure that: \begin{align*} \lim_{n\rightarrow \infty} \int_{-1}^{1}f_{k_n}(x)dx = \int_{-1}^{1}f(x)dx \in [0,1] \end{align*}

therefore $f \in \mathfrak{F}$, and we conclude that $\mathfrak{F}$ is closed.

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  • $\begingroup$ As a reply to How can a (continuous) function with a bounded domain and a bounded integral not be bounded? there is $\int_{0}^{1}\log(x)\,dx$. $\endgroup$ – Jack D'Aurizio Oct 22 '17 at 23:44
  • $\begingroup$ @JackD'Aurizio but then there is discontinuity at 0. $\endgroup$ – migueldva Oct 22 '17 at 23:52
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    $\begingroup$ $\log(x)$, as a function defined on $x\in(0,1)$, is continuous. And $(0,1)$ is a bounded interval. $\endgroup$ – Jack D'Aurizio Oct 22 '17 at 23:55
  • $\begingroup$ for example en.wikipedia.org/wiki/Rayleigh_fading#/media/… and assume that the domain is in $[0,1]$ $\endgroup$ – Seyhmus Güngören Oct 22 '17 at 23:56
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Consider the sequence $(f^n(x))_{n\in\mathbb N}$ defined by

\begin{equation*} f^n(x)=\begin{cases} nx^n & \text{if $n$ is odd} \\ \begin{cases} -nx^n & x\leq 0\\ nx^n & x>0 \end{cases} & \text{if $n$ is even} \end{cases} \end{equation*} This is a sequence in $\mathfrak F$ since at each $n$ the integral of $f^n$ is $0$ being the function antisymmetric about the origin. Now just compute the norm.

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If $T:C[-1,1]\to \Bbb R$ is the (linear) map $T(f)=\int_{-1}^1 f(x)\,dx$, then $\mathfrak F\supseteq \ker T$. Now, clearly $\ker T$ is a non-zero vector subspace. Thus, it is not bounded.

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Hint: consider a function that is 0 for most of the interval, but is large for a small chunk of the interval.

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