(Question from Royden's Real Analysis) Let C be the space of all continuous real-valued functions on [0,1], equipped with the sup norm metric.

Let $F_n=${$\exists x_0 \in [0,1]$ s.t. $\forall x \in (x_0,1]$, $f(x)-f(x_0) \leq n(x-x_0)$}

Prove that $F_n$ is a closed subset of C.

Update: I seem to have disproved the question

Let $f(x)=2nx$

$f_k(x)=2nx$ for $x\in[0,1-\frac{1}{k}]$
$=n(x-1+\frac{1}{k})+2n(1-\frac{1}{k})$ for $x \in (1-\frac{1}{k},1)$
(basically a polygonal function joining segments of slope 2n and n respectively)

Then $f_k$ converges to $f$ in C, each of $f_k$ are in $F_n$ but $f$ is not in $F_n$


Let $(f_k)$ be a sequence in $F_n$ converging to a limit $f$. Let $x_k\in [0,1]$ be the point such that $f_k(x)-f_k(x_k)\le n(x-x_k)$ for all $x\in (x_k,1]$. We can assume that $(x_k)$ converges to a limit $x_0$, passing to a subsequence if necessary. We show that $f(x)-f(x_0)\le n(x-x_0)$ if $x\in (x_0,1]$. Let $\epsilon>0$ and $x\in (x_0,1]$ be arbitrary. Since $C$ is complete, $f$ is continuous. Choose $k$ such that the following holds: $x_k<x,|f(x_k)-f(x_0)|<\frac\epsilon4,\|f_k-f\|_\infty<\frac\epsilon 4$ and $|x_0-x_k|<\frac{\epsilon}{4n}$.

Applying the triangle inequality yields:$$\begin{align} f(x)-f(x_0)&=f(x)-f_k(x)+f_k(x)-f_k(x_k)+f_k(x_k)-f(x_k)+f(x_k)-f(x_0)\\ &\le \frac\epsilon4+n(x-x_k)+\frac\epsilon4+\frac\epsilon4\\ &=n(x-x_0)+n(x_0-x_k)+\frac{3\epsilon}4\\ &\le n(x-x_0)+\frac\epsilon4+\frac{3\epsilon}4=n(x-x_0)+\epsilon. \end{align}$$ Since $\epsilon$ was arbitrary, the claim follows.


Hint : You can consider a Cauchy Sequence $\{f_n^i\}$ in $F_n$. Then since $F_n$ is a subset of C, so the sequence is Cauchy in C and since C is complete the sequence converges in C to some point $x$. Now show that $x\in F_n$.

So you have effectively proved that $F_n$ is complete and we have a fundamental result which states that a subset of a complete space is closed $\iff$ it is complete.

  • $\begingroup$ Your answer leaves out the "hard part" (proving that the limit is in $F_n$), but it gets the reader to that point, so +1. It might be clearer to call the limit $f$. $\endgroup$ – Ian Mar 14 '15 at 14:28
  • $\begingroup$ @Ian, Right. Atleast the reader will get some familiarity in going about abstract proofs and the "hard part" can be worked out so as to develop rigour. Cheers !! $\endgroup$ – creative Mar 14 '15 at 14:35
  • 1
    $\begingroup$ I am precisely stuck at the 'hard part'... $\endgroup$ – Chi Cheuk Tsang Mar 14 '15 at 14:39
  • $\begingroup$ @Tsang, I feel triangular inequality should work out ie $|f(x)-f(x_0)|\le |f(x)-f_n^i(x)|+|f_n^i-f(x_0)|$. Just check your luck. $\endgroup$ – creative Mar 14 '15 at 14:43

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