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.

  • 2
    $\begingroup$ What are your thoughts on this problem? Have you tried anything? $\endgroup$ – T. Eskin Mar 20 '13 at 8:21
  • $\begingroup$ 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. $\endgroup$ – John Mar 20 '13 at 8:28
  • $\begingroup$ Well have you shown it is not closed? to show boudned: what is every function bounded by, pointwise? $\endgroup$ – Lost1 Mar 20 '13 at 11:56
  • $\begingroup$ 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? $\endgroup$ – John Mar 20 '13 at 17:44
  • $\begingroup$ 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... $\endgroup$ – Julien 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 $P$.

  • $\begingroup$ 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? $\endgroup$ – John Mar 21 '13 at 6:15
  • $\begingroup$ 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 $x$. $\endgroup$ – Julien Mar 21 '13 at 6:18
  • $\begingroup$ 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. $\endgroup$ – John Mar 21 '13 at 6:24
  • $\begingroup$ 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. $\endgroup$ – Julien Mar 21 '13 at 6:27
  • $\begingroup$ Sorry for the basic question but what does := mean? $\endgroup$ – John Mar 21 '13 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.