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Define the subset of $C^0[0,1]$ to be: $P = \{F(x) = \int_0^x f(t) \, dt : f \in C^0[0,1], \|f\|_\infty \le 1\}$

1) Show that $P$ is not closed.

2) Show that $P$ is bounded and equicontinuous (using the infinity norm)

3) Show that the functional $J: C^0[0,1] \to R$ given by $J(F) = \int_0^1 F(x)\, dx$ achieves its max value on $P$.

Thanks in advance for your help and explanations. Thus far I have been working on the first two parts and have an intuition for why they are true, but I am struggling to construct a formal proof. I see that the functions are all differentiable and have been thinking about pointwise convergence. For part 2, I have been considering using the mean value theorem. Any help would be greatly appreciated.

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What are your thoughts on this problem? Have you tried anything? – T. Eskin Mar 20 '13 at 8:21
I've worked on the first two parts and haven't really gotten anywhere. It appears to be bounded but I'm not sure how to formally prove this. I see that these functions are all continuous and don't vary much within a given neighborhood so equicontinuity makes sense. Again, I am struggling with a formal proof. – John Mar 20 '13 at 8:28
Well have you shown it is not closed? to show boudned: what is every function bounded by, pointwise? – Lost1 Mar 20 '13 at 11:56
I have not been able to prove that it is not closed. I can't seem to come up with a limit point not in the set. Is it bounded by 1? – John Mar 20 '13 at 17:44
1-Your functions are differentiable. Is every uniform limit of differentiable functions differentiable? The polynomials are dense and differentiable. If this was true, every continuous function would be differentiable... – 1015 Mar 21 '13 at 1:48

1) Observe that every function in $P$ is differentiable. To show that $P$ is not closed, it suffices to construct a sequence in $P$ which converges uniformly to a non differentiable function $F$. For every $n\geq 1$, consider the continuous function $f_n$ defined by $1$ on $[0,1/2-1/n]$, $-1$ on $[1/2+1/n,1]$, and by an affine piece connecting these two portions. Then $\|f_n\|_\infty$, so $F_n(x):=\int_0^xf_n(t)dt$ belongs to $P$. I claim that $F_n$ converges to $F(x)=x$ on $[0,1/2]$ and $F(x)=1-x$ on $[1/2,1]$. This is straighforward by the dominated convergence theorem. Since $F_n$ is easily seen to be uniformly Cauchy, uniform convergence to $F$ follows. Of course, $F$ is not differentiable at $1/2$. So $P$ is not closed.

2) This is easier. First $$ |F(x)|\leq \int_0^x|f(t)|dt\leq \int_0^x\|f\|_\infty dt=\|f\|_\infty x\leq \|f\|_\infty\leq 1\qquad\forall x\in[0,1] $$ for all $F$ in $P$. Taking the sup, this yields $\|F\|_\infty\leq 1$ for all $F$ in $P$, so $P$ is bounded.

Second take $F$ in $P$ and check $$ |F(x)-F(x_0)|=\lvert\int_{x_0}^x f(t)dt\rvert\leq\lvert \int_{x_0}^x |f(t)|dt\rvert\leq |x-x_0|\|f\|_\infty\leq |x-x_0|. $$ It follows that $P$ is (uniformly) equicontinuous.

3) Note that for every $F$ in $P$ $$ |J(F)|\leq\int_0^1\left(\int_0^x|f(t)|dt\right)dx\leq\|f\|_\infty\int_0^1xdx\leq \frac{1}{2}. $$ And this is achieved for $f(t)=1$ and $F(x)=\int_0^x1dt=x$ in $P$.

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Thank you for your explanation. I actually had something similar for part 2 and your solution makes a lot of sense. I am still having a bit of trouble understanding part 1). How do you show that Fn is uniformly cauchy? – John Mar 21 '13 at 6:15
The functions $f_n$ and $f_m$ are equal, except on an interval of length $2/n$ (assuming $m\geq n$). And their difference is bounded by $2$. So $|F_n(x)-F_m(x)|\leq\int_0x|f_n-f_m|\leq 2\cdot 2/n$ for all $m\geq n$ and all $x$. – 1015 Mar 21 '13 at 6:18
Sorry I'm still having a bit of trouble understanding your solution. I don't quite get the justification for how the fn you constructed is an element of P. Also, is the idea that the function F it converges to is not differentiable at 1/2? Thank you. – John Mar 21 '13 at 6:24
Correct, $F$ is not differentiable at $1/2$. Now since $|f_n|\leq 1$, setting $F_n(x):=\int_0^xf_n$ defines an element of $P$ by definition. – 1015 Mar 21 '13 at 6:27
Sorry for the basic question but what does := mean? – John Mar 21 '13 at 6:30

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