Let $f \in C[a,b]$ define $\|f\| _{1} =\int_{a}^{b} |f|.$
(b)Show that there is no number $l$ such that$\|f\|_{\max} \leq l \|f\| _{1}, \forall f \in C[a,b].$
Could anyone help me, please?
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Without loss of generality, let $[a,b]=[-1,1]$. Consider $f_{n}(x)=-1$ for $-1\leq x\leq -1/n$ and $f_{n}(x)=x$ for $-1/n\leq x\leq 1/n$, and $f_{n}(x)=1$ for $1/n\leq x\leq 1$.
Argue that $\{f_{n}\}$ is Cauchy in $L^{1}$, then by the assumption, we have $\{f_{n}\}$ being Cauchy in $\sup$ norm, but the limit point is not a continuous function, a contradiction.
user284331
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Let $f_n(x)=n(\frac 1 n-x) $ for $0 \leq x\leq \frac 1n$ and $f_n(x)=0$ for $x> \frac 1 n$. Then the sup norm tends to $\infty$ but the $L^{1}$ norm is bounded.
In fact the sup norm of $f_n$ is $1$ and $\int_0^{1} |f_n(t)| dt=\frac 1 {2n}$.
Note: you can easily transform this example to get a similar sequence on any interval $[a,b]$: define $g_n(x)=f_n(\frac {x-a} {b-a})$.
Kavi Rama Murthy
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1how did you think about this example? – Emptymind Nov 17 '19 at 06:04
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so mine will not work? – Emptymind Nov 17 '19 at 06:04
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1@Emptymind Your example works only fro certain values of $a$ and $b$. It works if $|a|<1$ and $|b| <1$. – Kavi Rama Murthy Nov 17 '19 at 06:08
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How did you think about this example? – Emptymind Nov 17 '19 at 12:17
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how I can transform it? – Emptymind Nov 17 '19 at 12:20
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1I have given the construction for any $[a,b]$ now. The idea for $(f_n)$ is to make $|f_n|1$ bounded but $|f_n|{max}$ go to $\infty$. For the second property it is enough if $f_n(0) \to \infty$. Now maker $f_n$ $0$ outside some small interval around $0$ to get the first property. @Emptymind – Kavi Rama Murthy Nov 17 '19 at 12:22
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I think the supremum norm is 1 and not $n$ in your example on [0,1]. correct? – Emptymind Nov 19 '19 at 07:03
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@Emptymind You are right. The supremum is $1$. – Kavi Rama Murthy Nov 19 '19 at 07:21
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Now for the adjusted sequence of functions the supremum norm is $1 + \frac{a n}{b-a}$ while the other norm is $\frac{1}{n} - \frac{1}{2n(b-a)} + \frac{a}{b-a}$ which is bounded by 1..... are my calculations correct? – Emptymind Nov 19 '19 at 07:30
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Check your calculations again. The supremum of $g_n$ is $1$ and $\int_a^{b} g_n(x)dx=(b-a)\int_0^{1} f_n(y)dy=\frac {b-a} {2n}$. @Emptymind – Kavi Rama Murthy Nov 19 '19 at 07:41
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$y = \frac {x - a}{b-a}$? – Emptymind Nov 19 '19 at 07:46
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Yes. That is the substitution you have to make. – Kavi Rama Murthy Nov 19 '19 at 07:47
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And what are the bounds for $x$ in your new function $g$? are they still $0 \leq x \leq (1/n)$? – Emptymind Nov 19 '19 at 07:54
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1@Emptymind There is no need to worry about that. Once you know $|f_n|_{\infty}$ and $\int |f_n|$ simply make the substitution to find the corresponding numbers for $g_n$. Don't go back to the definition of $f_n$. – Kavi Rama Murthy Nov 19 '19 at 07:57
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but the integral of |f_{n}| = (1/2n) how can I make substitution in this ? – Emptymind Nov 19 '19 at 08:11