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Prove or contradict that the following norms are equivalent:

$$ \begin{align*} \|f\|_1 &= \int_0^1 |f| \; d\lambda+ \max \; |f'|, \\ \|f\|_2 &=\max \; |f|+\max \; |f'|, \end{align*}$$

for all $f\in C_1 ([0,1])$.

I did prove it for nonnegative functions (with mean value theorem) but I don't know how to continue from there, I would be glad to get some help. Thanks!

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why do we have max|f|≤max|f′| ? –  user18217 Dec 5 '11 at 17:13
    
Can you prove inequality in one direction or the other? –  GEdgar Dec 5 '11 at 18:48
    
of course ||f||1≤ ||f||2 , that's easy. –  user18217 Dec 5 '11 at 19:30

2 Answers 2

up vote 1 down vote accepted

Let $f \in C_1([0,1])$

$$\int |f| d\lambda \leq \int \max |f|\, d\lambda, \text{ so }\|f\|_1 \leq \|f\|_2.$$

Now, since $f'$ is continuous on closed interval $[0,1]$ it is bounded. Therefore for each $x \in [0,1]$ we derive

$$ \begin{align} \|f\|_2 &= \max|f| + \max|f'|\newline &\leq |f(x)| + \max|f'| + \max|f'|\newline &\leq |f(x)| + 2\|f\|_1 - 2\int |f|d\lambda\newline &\leq |f(x)| + 2\|f\|_1. \end{align} $$

However we know that $\exists_{x_0} |f(x_0)| \leq \int |f|\, d\lambda$, so we have $|f(x_0)| \leq \|f\|_1$ and finally

$$\|f_2\| \leq 3\|f\|_1.$$

Why $\max|f| \leq |f(x)| + \max |f'|$?

Let $x,y \in [0,1]$. Then $\exists_{c \in [0,1]}$ such that $$\big| |f(x)| - |f(y)|\big| \leq |f(x) - f(y)| \leq |f'(c)| \leq \max |f'|$$ by the mean value theorem.

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can you please explain the first inequality? –  user18217 Dec 5 '11 at 19:24
    
there should be 1 instead of 1/2, I've just edited and given explanation at the end. –  Łukasz Kidziński Dec 5 '11 at 19:36
    
now it's clear, thanks! –  user18217 Dec 5 '11 at 19:59

Obviously $\|f\|_1 \leq \| f \|_2$. This tells us that the identity operator $(C^1[0,1],\| \cdot \|_2) \to (C^1[0,1], \| \cdot \|_1 )$ is continuous, if we can prove this is an homeomorphism we're done. We know that it's a bijective, bounded linear map so, by the open mapping theorem, it suffices to check that $\| \cdot \|_1$ induces a complete metric.

Take a Cauchy sequence in $(C^1[0,1], \| \cdot \|_1 )$ then, in particular, $f_n'$ is a Cauchy sequence in $C[0,1]$ so $f_n'\to g\in C[0,1]$ uniformly. Since $f_n$ is Cauchy in $L^1$ we have $f_n \to f\in L^1[0,1]$. It's standard that there exists a subsequence of the $f_n$ that converges pointwise a.e. in particular it does so at a point, but the sequence of derivatives converges uniformly, so that $f\in C^1[0,1]$ and $g=f'$.

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