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Consider the sequence of functions

\begin{equation} f_n(x) = x^n, \quad x \in [0, 1]. \end{equation}

Is this sequence Cauchy in $(C[0, 1], ||\cdot||_{\infty})$?

The pointwise limit is not continuous because we have

\begin{equation} f(x) := \lim_{n \to \infty} f_n(x) = \begin{cases} 0 & \mbox{if $x \in [0, 1)$} \\ 1 & \mbox{if $x = 1$} \end{cases} \end{equation}

Morever we have

\begin{align} ||x^n - x^m||_{\infty} &= \sup_{x \in [0, 1]} |x^n - x^m| \\ &\leq \sup_{x \in (0, 1)} |x^n| + |x^m| \\ &\leq 2 \varepsilon. \end{align}

The first inequality uses the triangle inequality and the fact that $x^n - x^m = 0$ for $x \in \{0, 1\}$. The second inequality uses the fact that for $x^n \to 0$ for $x \in (0, 1)$ so we can find an $N$ such that $x^n < \varepsilon$ for any $n > N$.

Thus $f_n$ is Cauchy in $(C[0, 1], ||\cdot||_{\infty})$, which is complete. Hence $f_n \to f \in (C[0, 1], ||\cdot||_{\infty})$. But we already showed that $f$ is not continuous. Contradiction.

Can anyone tell me where things are going wrong?

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4 Answers 4

up vote 2 down vote accepted

The inequality $$|x^n|+|x^m|\le 2\varepsilon$$ is true for any fixed $x\in(0,1)$ and for any given $\varepsilon>0$, starting from some $N$ (i.e. for all $n,m\ge N$).

But the number $N$ depends on $x$. (You are using the fact that $x^n\to 0$ for $x\in (0,1)$, but this is only pointwise convergence, not uniform.)

So you don't get $$\sup_{x\in(0,1)}|x^n|+|x^m|\le 2\varepsilon$$ for $m,n>N$ from the above inequality. (In order to get this you would need to have the same $N$ for each $x\in(0,1)$.)

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We have $\sup_{x\in (0,1)}|x|^n=1$, so the argument in the OP doesn't work. An other reason it that the sequence $\{f_n\}$ is not Cauchy for the supremum norm. Indeed, if it was the case, then $\sup_{x\in (0,1)}|x^{2n}-x^n|$ would be $0$. Let $g_n(x):=x^n-x^{2n}=x^n(1-x^n)$. If $x^n=1/2$, $x=2^{-1/n}$ so $\sup_{x\in [0,1]}|x^{2n}-x^n|\geq \frac 14$ (it's actually equal but we don't need that).

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No, the sequence is not Cauchy $(C[0, 1], ||\cdot||_{\infty})$. A formal proof is given below.

Proof. By definition if $x^n$ is Cauchy, then for all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that

\begin{equation} ||x^n - x^m||_{\infty} < \varepsilon \end{equation} whenever $n, m \geq N$.

Fix $\varepsilon > 0$. Suppose for a contradiction that such an $N$ exists. Then in particular we can set $n = N$ and it must be the case that for all $m \geq N$

\begin{equation} ||x^N - x^m||_{\infty} < \varepsilon. \end{equation}

But as $x^N$ is continuous and equal to $1$ when $x = 1$. Thus for any $\eta > 0$ we can find a $\delta > 0$ such that $|1 - x^N| < \eta/2$ whenever $|1 - x| < \delta$. But this means $x^N > 1 - \eta/2$. Let $x_0$ be arbitrary in $(1-\delta, 1)$. At $x_0$ there exists $m$ such that $x_0^m < \eta/2$. Thus at $x_0$ we have

\begin{equation} |x_0^N - x_0^m| = x_0^N - x_0^m \geq 1 - \eta/2 - \eta/2 = 1 - \eta. \end{equation}

Choose $\eta$ so that $1 - \eta > \varepsilon$. Then

\begin{equation} ||x^n - x^m||_{\infty} = \sup_{x \in [0, 1]} |x^n - x^m| \geq |x_0^n - x_0^m| \geq \varepsilon. \end{equation}

A contradiction. Thus $x^n$ is not Cauchy in $(C[0, 1], ||\cdot||_{\infty})$.

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I have thought of a simpler answer, which I believe to be correct, it plays on the assumption of cauchy, and then letting $m=N$, and $n=N+1$.

Then: $\|x^m-x^n\|_\infty= ||x^{N}-x^{N+1}\|_\infty = \|x^{N}(1-x)\|_\infty $

Now:

$ ||x^{N}(1-x)\|_\infty = \sup_{x\in [0,1]}|x^{N}| |(1-x)| \geq \frac{1}{2}^N\times\frac{1}{2} = \frac{1}{2}^{N+1} $

Now let $ \varepsilon =\frac{1}{4}^{N+1} $, so $\|x^N-x^{N+1}\|_\infty\geq \frac{1}{2}^{N+1} \gt \frac{1}{4}^{N+1} = \varepsilon $

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