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Let: $$f_n(x) = \begin{cases} 1 & -n \le x \le n \\ 0 & \text{ otherwise} \end{cases}$$ Prove that $\|f_n\|_∞ = 1$, but $\|f_n\|_2 \to \infty$ as $n \to \infty$ and that there is no constant $C$ such that $\|f\|_2 \le C\|f\|_\infty$ for all functions $f$.

My attempt:

I know that for $\|f_n\|_2$ it goes to $\infty$ since the integral will equal to $x$ from the limits $-n \to n$ and if $n = \infty$ we have that $\|f_n\|_2 = \infty$. But why does $\|f_n\|_\infty = 1$? Also, since they both don't approach the same limit it means there is no constant $C$?

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I improved the $\LaTeX$ markup in your question. Please double-check to make sure this is what you meant. –  Ayman Hourieh Oct 1 '12 at 22:21
    
I think the tag Functional-Analysis might also be appropriate since this is where I encountered problems like this in my math education. –  Carl Morris Oct 1 '12 at 22:25

2 Answers 2

up vote 2 down vote accepted

The definition of $M=||f||_\infty$ is the essential supremum of $f$. In other words, it is the the smallest number $M$ such that $f(x)\leq M$ for $x$ almost everywhere (in your case with respect to the Lebesgue measure, most likely). In your case the essential supremum is the same as the regular supremum, because your function is well behaved. The largest value that $f_n$ can take for any $n$ is 1, so that is the value of $||f_n||_\infty$.

For the second part, because $||f_n||_2\rightarrow\infty$ as $n\rightarrow\infty$, for any fixed $C$, I can choose $n$ large enough such that $||f_n||_2>C=C||f||_\infty$. Therefore, no such $C$ could ever satisfy $||f_n||_2\leq C||f||_\infty$ for all $n$.

EDIT:

In case the supremum norm is defined by $||f||_\infty=\lim_{p\rightarrow\infty}||f||_p$, you can use the following calculation. $$ ||f_n||_p=\left(\int_{-n}^n 1^p dx\right)^{1/p} = (2n)^{1/p} $$ This right hand side converges to 1 for all $n$ as $p\rightarrow\infty$, hence $||f_n||_\infty=1$ for all $n$.

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I dont understand your first part because we haven't learned supremum yet and thank you very much for the second part! –  diimension Oct 1 '12 at 22:43

Check your definition of $\|f_n\|_\infty$. If you've defined it as the essential supremum, it's not hard to see that $f^{-1}(a,\infty) = \emptyset$ if $a > 1$ and $f^{-1}(a,\infty) \supset [-n,n]$ for $a \leq 1$ and so you can show $\|f\|_\infty = 1$. On the other hand, if your definition is something like $\lim_{\ p\rightarrow \infty} \|f_n\|_p$, then calculating the value of the integral isn't difficult. Remember, the $\infty$ norm is supposed to act like a "maximum" or "supremum" of the function's absolute value, which is obviously 1 here.

As for the constant, it looks like you've got it. If there was such a constant, it would hold for your $f_n$, which it obviously doesn't.

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I meant to put the lim because we haven't learned supremum yet and how would you calculate the value. I got that it was approaching infinity since the integral is x. ? –  diimension Oct 1 '12 at 22:44
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You would take $\int |f_n|^p$, which is $2n$. So $\|f_n\|_p = (2n)^\frac{1}{p}$. Now take the limit as $p \rightarrow \infty$. –  Zach L. Oct 1 '12 at 22:55

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