0

Consider the sequence $f_n(x)=x^n \subset C[0,1]$ (the space of continous functions on the interval $[0,1]$) and the following norms:

$$\lvert \lvert f\lvert \lvert _\infty=\max\left\{ \lvert(f(x) \lvert x \in [0,1] \right\} \hspace{1cm} \lvert \lvert f\lvert \lvert _1=\int_0^1 \lvert f(x)\lvert dx$$

Check if $f_n$ is convergent in $C[0,1], \lvert \lvert\cdot\lvert \lvert_1$ and/or $C[0,1], \lvert \lvert\cdot\lvert \lvert_\infty$.

Problem: Since it is not specified, I am not quite sure what limit point / limit function I am supposed to take here. I know that a sequence $(x_n)_{n\in \mathbb N}$ converges to $x$ if:

$$\lim_{n \to \infty} \lvert \lvert\ x_n-x \lvert \lvert=0$$

I was thinking of setting $x$ to the limit point of $\lim_{n\to\infty}x^n$, but the sequence has different limit points for $\lvert x\lvert<1$ and for $\lvert x\lvert=1$. Do I need to show that it converges (or doesn't) converge to both limit points or do I just take the "largest" limit $x=1$?

Nullspace
  • 979
  • How about checking Cauchy criterion? – Riemann Nov 20 '22 at 13:03
  • Whether the sequence of functions converges is a different thing from whether the sequence of values at a single point $x$ converges. – aschepler Nov 20 '22 at 13:04
  • the usual candidate is the pointwise limit, like you've done, but keep in mind that this is true up to a "negligible set" of points, that typically means "up to a finite number of points" – Exodd Nov 20 '22 at 13:07
  • @aschepler I am a bit confused. Have I misunderstood the question? – Nullspace Nov 20 '22 at 13:09
  • @Exodd But aren't there two pointwise limits for $\lvert x \lvert<1$ and $\lvert x \lvert =1$ One is zero and the other is $1$? – Nullspace Nov 20 '22 at 13:12
  • @Nullspace you are looking for a limit function $g(x)$. You have observed that $x^n\to 0$ if $x\in [0,1)$ and $x^n\to 1$ if $x=1$. So you can define $g(x)$ like this: $g(x) = 0$ if $x\in [0,1)$ and $g(x)=1$ if $x=1$ – Exodd Nov 20 '22 at 13:15
  • Maybe a little. The "limit point / limit function" piece is key - limit function is important here. To show $f_n$ is convergent in a given metric space, we need to give the limit function it approaches. To show $f_n$ is not convergent, we need to show it can't approach any particular limit function at all. – aschepler Nov 20 '22 at 13:16
  • @Exodd Except that piecewise limit function is not an element of $C[0,1]$. – aschepler Nov 20 '22 at 13:17
  • @aschepler Would the limit functions in this case be $\hat{f}(x)=1$ and $\hat{f}(x)=0$ (similar to what Exodd commented)? – Nullspace Nov 20 '22 at 13:22
  • @aschepler Ah I just realized what you meant by your comment that the limit function I have chosen is not continuous and therefore not $\in \mathcal C[0,1]$. I still don't understand what limit function I should be chosing instead though. – Nullspace Nov 20 '22 at 13:33
  • Well, do you think $f_n \to \hat f$ in either case? This means $\lim_{n \to \infty} |f_n-\hat f|=0$. – aschepler Nov 20 '22 at 13:34
  • @aschepler. Taking the infinity norm: $\vert \vert \cdot \vert \vert_\infty$, I would have for $x=1$ and the corresponding limiting function $\hat{f}=1$: $\lim_{n \to \infty}=\vert \vert f_n-\hat{f} \vert \vert = \lim_{n \to \infty } \max { x^n-1}=0$. For $0<x<1$ we have $\lim_{n \to \infty } \max { x^n-0}=?$. I am not sure about the maximum of the last statement. – Nullspace Nov 20 '22 at 13:55
  • @aschepler But I belive the last statement $\lim_{n \to \infty} \max { x^n }$ is also zero and therefore the sequence converges to both limiting functions. – Nullspace Nov 20 '22 at 14:02
  • If $f_n$ is convergent, it converges to one single limit function, not one limit function for some $x$ values and a different function for other $x$ values. – aschepler Nov 20 '22 at 14:05
  • @aschepler Okay that helps a lot. That would mean in this case, since $lim_{n \to \infty} f_n=0$ for $0<x<1$ and $lim_{n \to \infty} f_n=1$ for $x=1$, $f_n$ does NOT converge to a single limit function and is therefor not convergent on $\mathcal C[0,1]$ – Nullspace Nov 20 '22 at 14:09
  • @aschepler In case my reasoning is correct that $\lim_{n \to \infty} \max { x^n}=0$ (still not entirely sure if that is the case) – Nullspace Nov 20 '22 at 14:16

1 Answers1

1

First let's look at the infinity norm, $\| g \|_\infty = \max_{x\in[0,1]} |g(x)|$.

Claim: In the $C[0,1], \|\cdot\|_\infty$ metric space, if a sequence of functions $(g_n)$ converges to $g$, then at each $x \in [0,1]$, the sequence $g_n(x)$ converges to $g(x)$.

Proof: Let $g_n \in C[0,1]$ be a sequence of functions which converges to a function $g \in C[0,1]$, and let $x$ be any arbitrary point in $[0,1]$. Given a real $\epsilon > 0$, by convergence of $(g_n)$ to $g$, there is a natural $N$ such that

$$ n > N \implies \|g_n - g\|_\infty < \epsilon \iff \max_{t\in[0,1]} |g_n(t)-g(t)| < \epsilon $$

By the definition of $\max$, $|g_n(x)-g(x)| \leq \max_{t\in[0,1]}|g_n(t)-g(t)| < \epsilon$. Since this $N$ exists for every $\epsilon > 0$, this has proved that $\lim_{n \to \infty} g_n(x) = g(x)$.

So if $(f_n)$ converges to a function $f$, then for every $x$ we must have $f(x) = \lim_{n \to \infty} f_n(x)$, and $f$ must be the function

$$ f(x) = \begin{cases} 0 & x<1 \\ 1 & x=1 \end{cases} $$ Since this function is not continuous, it's not possible that $(f_n)$ converges to a function in $C[0,1]$. In other words, $(f_n)$ is not convergent.

Now look at the $1$-norm, $\|g\|_1 = \int_0^1 g(x)\, dx$. We know that the value of an integral does not change if the value of the integrand function changes at a finite number of points. So although the pointwise limit of $(f_n)$ is not continuous, there's a similar function which is "mostly equal to" the pointwise limit: the constant function $f(x)=0$. To check whether $(f_n) \to f$, we can explicitly evaluate the limit of the difference:

$$ \begin{align*} \lim_{n \to \infty} \|f_n-f\| &= \lim_{n \to \infty} \int_0^1 (f_n(x)-f(x))\, dx \\ &= \lim_{n \to \infty} \int_0^1 (x^n-0)\, dx \\ &= \lim_{n \to \infty} \left(\frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1}\right) \\ &= \lim_{n \to \infty} \frac{1}{n+1} \\ \lim_{n \to \infty} \|f_n-f\| &= 0 \end{align*} $$

So yes, in the $C[0,1],\|\cdot\|_1$ metric space, $(f_n)$ converges to the zero function.

aschepler
  • 9,449
  • Thank you very much for your answer. The first part is percectly clear. I don't understand the sentence "We know that the value of an integral does not change if the value of the integrand function changes at a finite number of points." Is that because the integral i.e. the "area" at just the point $x=1$ is zero? – Nullspace Nov 20 '22 at 16:07
  • That's the intuitive idea, yes. See for example https://math.stackexchange.com/questions/102844/how-do-i-prove-that-a-function-with-a-finite-number-of-discontinuities-is-rieman . This isn't technically part of the proof, just an explanation of why we would guess the limit function is the zero function. – aschepler Nov 20 '22 at 21:21