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In the Banach space $C[0,1]$ consider the subspace

$M=\lbrace g \in C[0,1]: \int_{0}^{1}g(t)dt=0 \rbrace $

Show that $M$ is closed in $C[0,1]$ and calculate the quotient norm $(\|f+M \|)$ where $f(t)=\sin(\pi t)$ for all $t \in [0,1]$.

Probably it's easier than I think but I don't know how do this, it is obvious to see that $M$ is closed but how can I show this? And what about the norm? No idea... please help me

Martin Sleziak
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Maggie94
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    Can you show that $g\mapsto \int_0^1 g(t) dt$ is a linear continuous functional on $C[0,1]$? From this you should get that $M$ is a closed subspace. – Martin Sleziak Dec 07 '18 at 12:32
  • What function is $sen(\pi t)$? A natural guess is that you mean $\sin\pi t$. I see that in some languages it is called seno and even denoted by $\operatorname{sen}$. – Martin Sleziak Dec 07 '18 at 12:35
  • I suppose it's in the infinity norm right? Or you want to do it with open sets? – mmcrjx Dec 07 '18 at 12:39
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    Yes, it is the sin function, I corrected it, and yes it is the infinity norm – Maggie94 Dec 07 '18 at 13:52
  • And what about the norm? – Maggie94 Dec 07 '18 at 13:54
  • Yes I think I can show that the functional that given $g$ gives me the integral is linear and continuous, and why it is enought to show that the subspace is closed? – Maggie94 Dec 07 '18 at 14:19
  • @Maggie94 The idea how to use this to show that the subspace is closed is using the fact that $M$ is exactly the kernel of this functional. (Which is already mentioned in the answer.) At the moment I do not see an easy approach to calculate the norm, I have mentioned this in chat, maybe somebody will notice the question there. (Unfortunately, you'll need at least 20 reputation points to be able to talk in chat.) – Martin Sleziak Dec 08 '18 at 10:22
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    @Maggie94 I will mention that your question has received several close votes and at the moment it is still in the close votes review queue. It is most likely that without improvements the question will be closed and eventually deleted. It might help if you read the part related to context in the FAQ post How to ask a good question. You have probably seen this link in some of your previous posts that were put on hold. – Martin Sleziak Dec 08 '18 at 11:26
  • Here are some of the reasons why it might be useful to improve the question (or at least ask questions in the future where at least some context is included): 1) If you get some comments or answers, the work of the people who wrote them is not wasted. 2) It improves the chances that you actually get some answers. 3) It makes the post more useful for others. 4) It helps you to avoid question ban - if many of your questions get closed (downvoted, deleted), you might be banned from asking more questions. – Martin Sleziak Dec 08 '18 at 11:30
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    To include some examples of what things might be added as context in this question: 1) Where does the question come from? (If it is from some book, you might include full citation.) 2) You could at least include the definition of the quotient norm $|f+M|_{X/M}$ so that the answerers see that you know what you are trying to calculate. 3) You could have included the proof that the functional in the first comment is indeed continuous (at least after it was mentioned that this functional is useful for your problem). – Martin Sleziak Dec 08 '18 at 11:32

2 Answers2

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I am just expanding @Martin Sleziak's suggestion.

We consider a functional $L:g\to \int_{0}^{1}g(t)dt$. And prove that the map $L$ is linear and continuous(which is trivial). Now the kernel of $L$ is $M$ according to definition of $M$. And The kernel of a continuous linear operator is a closed subspace(the range $\mathbb{R}$ is finite dimensional). So M is closed.

Now for the quotient norm We have to calculate $\left\lVert[\sin(\pi t)]\right\rVert_{C[0,1]/M}$ which is given by $\inf\limits_{x\in M}\left\lVert\sin(\pi t)-m\right\rVert_{\infty}$. We know $g \equiv 0 \in M$. Therefore $\left\lVert[\sin(\pi t)]\right\rVert_{C[0,1]/M} \leq \left\lVert\sin(\pi t)\right\rVert_{\infty}=1$.

Community
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mmcrjx
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    Maybe I misunderstood something, but it seems that $\sup\limits_{m\in M}|m|\infty=+\infty$ and thus $\inf\limits{m\in M}(|\sin\pi t|-|m|)=-\infty$. – Martin Sleziak Dec 08 '18 at 10:00
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    I also think that the norm must less than one. For example, if we take $f(t)=\sin\pi t$ and $g(t)=-\frac16\sin(3\pi t)$ than $|f(t)-g(t)|_\infty<1$. Instead of detailed computation I'll add link to WA and also to plot of the graphs. Finding the actual infimum seems to be a more difficult problem. – Martin Sleziak Dec 08 '18 at 10:02
  • Yeah the greater than proof is fishy now.. I will check it again. But norm is a non negative quantity so cannot be $-\infty$ for sure. So its in $(0,1]$ – mmcrjx Dec 08 '18 at 10:09
  • Yes, that's true. Basically all I am saying is that the way it is currently written, this estimate only gives $|f|_{X/M}\ge-\infty$, which is nothing new - as you correctly say. I left a few comments about this question in the functional analysis chatroom - maybe somebody who has an idea how to continue will notice that it was mentioned there. (Of course, you're welcome in that room, too.) – Martin Sleziak Dec 08 '18 at 10:14
  • I am new to this idea of chat rooms but I will have a look .. Anyway, that kernel idea was brilliant .. #kurnis – mmcrjx Dec 08 '18 at 10:17
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Let us denote $X=C[0,1]$. As mentioned in the comments and in the other answer, if we check that $F\colon X\to\mathbb R$ $$F\colon g\mapsto\int_0^1 g(t)dt$$ is a linear continuous functional, then $M=\ker F=F^{-1}(\{0\})$ and the continuity of $F$ implies that $M$ is closed.

We are also interested in the norm $\|f+M\|_{X/M}$ for $f(t)=\sin(\pi t)$. This norm can be expressed in several equivalent forms $$\|f+M\|_{X/M} = \inf_{g\in M} \|f-g\|_{\infty} = \inf_{h\in f+M} \|h\|_\infty.$$ You may notice that $h\in f+M$ is equivalent to $F(h)=F(f)$, this follows again from the fact that $M=\ker F$.

Specifically, for $f=\sin\pi t$ we get $F(f)=\frac2\pi$. So our problem basically reduces to finding the infimum of $\|h\|_\infty$ when we are looking at the functions $h\in C[0,1]$ such that $$\int_0^1 h(t) dt = \frac2\pi.$$

We can notice that if we choose the constant function $h_0(t)=\frac2\pi$ we have $h_0\in f+M$ and $\|h_0\|_\infty=\frac2\pi$.

We can also notice that for any function we get $$|F(h)| = \left|\int_0^1 h(t) dt\right| \le \int_0^1 |h(t)| dt \le \sup_{t\in[0,1]} |h(t)| = \|h\|_\infty.$$ This implies that $\|h\|_\infty\ge\frac2\pi$ for any $h\in f+M$.

So together we get that $$\|f+M\|_{X/M} = \frac2\pi.$$

Martin Sleziak
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