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Let $\hat{f}(t) := \int_\mathbb{R} f(x) e^{-2 \pi i xt} d\lambda(x)$

I quote from my lecture notes:

Equation (3.4)

$$ sup_{t \in \mathbb{R}} | \hat{f} (t)| \leq || f ||_1$$

Remark:

In the language of normed vector spaces the inequality (3.4) means that the Fourier transform map $L^1 (\mathbb{R}) \rightarrow C_b (\mathbb{R})$, $f \rightarrow \hat{f}$ is continuous and has norm $\leq 1$, where $C_b (\mathbb{R})$ is the space of bounded continuous functions on $\mathbb{R}$, which is a normed vector space with the norm $|| f ||_\infty := sup_{x \in \mathbb{R}}|f(x)|$.

My question: what is the norm on the vector space of functions from $L^1 (\mathbb{R})$ to $C_b (\mathbb{R})$? Is it something like $ || \hat{} || := sup_{f \in L^1} \frac{sup_{t \in \mathbb{R}} | \hat{f} (t)|}{|| f ||_1} $?

Many thanks for your help!

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Yes, exactly. See en.wikipedia.org/wiki/Operator_norm#Equivalent_definitions –  t.b. Feb 6 '11 at 13:55
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Dear Matt: Let $f$ be a positive function. Then $\hat{f}(0) = \int f = ||f||_1$. So the norm is not just $\leq 1$, but precisely 1. –  Akhil Mathew Feb 6 '11 at 14:15
    
@Theo: yay : ) thanks. Now I'm pleased about my good guess. –  Matt N. Feb 6 '11 at 19:47

1 Answer 1

up vote 2 down vote accepted

To expand on the Wikipedia-Link I gave as a comment: Let $(V,\Vert\cdot\Vert_{V})$ and $(W,\Vert\cdot\Vert_{W})$ be normed linear spaces. Then a standard basic result of functional analysis asserts that for linear operator $T: V \to W$ the following are equivalent:

  1. $T$ is continuous at $0$
  2. $T$ is continuous
  3. $T$ is uniformly continuous
  4. The norm of $T$ defined by \[ \|T\| = \sup_{0 \neq v \in V} \frac{\Vert Tv\Vert_{W}}{\Vert v \|_{V}} \lt \infty \] is finite.

To see this, note that $\|T\| \lt \infty$ implies that $\Vert Tv_{1} - Tv_{2} \Vert_{W} \leq \|T\|\, \Vert v_{1} - v_{2}\Vert_{V}$ so that $T$ is in fact Lipschitz continuous with Lipschitz constant $\|T\|$. Hence we have the obvious chain of implications $(4.) \Rightarrow (3.) \Rightarrow (2.) \Rightarrow (1.)$. It remains to prove that $(1.) \Rightarrow (4.)$. Suppose $(4.)$ fails. Then there exists for each $n$ an $v_{n}$ with $\Vert T v_{n}\Vert_{W} \geq n \Vert v_{n} \Vert_{V}$ by the definition of $\|T\|$. Put $x_{n} = \frac{v_{n}}{n\cdot\Vert v_{n}\Vert_{V}}$ and note that $\Vert x_{n} \Vert_{V} = \frac{1}{n} \to 0$ so that $x_{n} \to 0$. On the other hand, $\|Tx_{n}\|_{W} \geq 1$ and hence $T$ can't be continuous at $0$.


Applying this to the Fourier transform $\mathcal{F}: L^{1} \to C_{b}$ you get $\|\mathcal{F}\| = \sup_{0 \neq f \in L^{1}} \frac{\Vert \mathcal{F}(f) \Vert_{C_{b}}}{\Vert f\Vert_{L^{1}}}$, which is precisely your guess (up to the minor detail $f \neq 0$). Recall that the norm $\|g\|_{C_{b}} = \sup_{t \in \mathbb{R}} |g(t)|$. Now the lemma you quote clearly says that $\|\mathcal{F}\| \leq 1$.

Let me finish by mentioning that in fact the image of $L^{1}$ under the Fourier transform is contained in $C_{0} \subset C_{b}$, the space of functions vanishing at infinity. This is the content of the Riemann-Lebesgue lemma, see also the first paragraph of this answer.


Added: I forgot to say why $\|\mathcal{F}\| = 1$, but Akhil has explained this in his comment to your question.

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Thanks for this most excellent, clear answer. –  Matt N. Feb 6 '11 at 19:49

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