Lecture 12

Tempered distributions¶

Similar to $\mathcal D$, the Schwartz functions $\mathcal S$ has a topology determined by the semi-norms $\|\cdot\|_{k,l}$. A sequence $f_n\to f$ in $\mathcal S$ if and only if for any $k,l\geq 0$, $\|f_n - f\|_{k,l}\to 0$ when $n\to \infty$. Recall that the distributions are “dual objects”, i.e. continuous linear functionals, of $\mathcal D$. The dual object of $\mathcal S$ are tempered distributions.

Definition of tempered distributions¶

A tempered distribution is a linear functional $T:\mathcal S\to \mathbb{C}$ such that for some $C>0$ and $N\in \mathbb{Z}_+$,

Similarly, one can show that this definition is equivalent to $T$ being continuous linear functional on $\mathcal S$ in the sense that for every $f_n\to f$ in $\mathcal S$, $T(f_n)\to T(f)$. To verify this one can use exactly the same trick as the case of $\mathcal D'$.

We have $\mathcal S'\subset \mathcal D'$ because $\mathcal D\subset \mathcal S$. But by the following lemma, a tempered distribution is completely determined by its action on $\mathcal S$.

Lemma¶

$\mathscr{C}_c^{\infty}(\mathbb R)$ is dense in $\mathcal S$.

Proof. Let $\chi_{n,1}$ be the cut-off function constructed in previous lecture. Let $f_n(x) = \chi_{n,1}(x) \cdot f(x)$. Then $f_n\in \mathcal D$. We claim that $f_n \to f$ in $\mathcal S$, i.e. $\|f_n - f\|_{(k,l)} = \|(1 - \chi_{1,n})f\|_{k,l} \to 0$ for every $k,l$.

Note that

Then $\|(1-\chi_{1,n})f\|_{k,l} = \sup_{n\leq|x|\leq n+1}|x^k((1-\chi_{1,n})f)^{(l)}|\leq C'\sup_{n\leq |x|\leq n+1}\sum_{m = 0}^l|x^k f(x)^{(m)}|$, where $C'$ is a constant depending on derivatives of $\chi_{1,n}$ up to order $l$. This goes to zero when $n\to \infty$ because $f\in \mathcal S$. $\square$

Let $T$ be a tempered distribution. Suppose we only know $T(\varphi)$ for every $\varphi\in \mathcal D$. Then for $f\in \mathcal S$, from the lemma we can choose $\varphi_n\in \mathcal D$ such that $\varphi_n\to f$ in $\mathcal S$. Then since $T$ is a tempered distribution, $T$ is continuous, so $T(f) = \lim_{n\to \infty}T(\varphi_n)$. Note that one can show that the sequence $T(\varphi_n)$ is Cauchy so the limit exists.

Examples of tempered distribution¶

Functions¶

Any locally integrable function is a distribution of degree 0. Let $f$ be a function on $\mathbb{R}$, it determines a distribution $T_f$ such that $(T_f,\varphi) = \int f \varphi$ for every $\varphi \in \mathcal D$. Indeed, for every compact set $K$, every $\varphi\in \mathscr{C}_K^{\infty}(\mathbb{R})$, $|(T_f,\varphi)| = |\int f\varphi|\leq \sup_{x\in K}|\varphi(x)|\cdot \int_K |f|$.

But a function may not be tempered. For example, $e^{x^2}$ is not a tempered distribution. $(T_{e^{x^2}},e^{-x^2}) = \int e^{x^2}\cdot e^{ - x^2} = \int 1 = \infty$.

This example tells us that in order for a function to be tempered distribution, it need some restriction, though not strict, on its growth.

Proposition. $(1+|x|)^{-m}f$ is bounded for some $m$, then $T_f$ is a tempered distribution.

Proof. For $g\in \mathcal S$, $(T_f,g) = \int fg = \int (1+|x|)^{-m}(1+|x|)^m (1+|x|)^2 (1+|x|)^{-2}g f \leq \|g\|_{m+2}\int C (1+|x|)^{-2}<\infty$, where $C$ is an upper bound of $(1+|x|)^{-m}f$.

The δ-measure¶

The δ measure is a distribution, its action on $\mathcal D$ is given by $(\delta,\varphi) = \varphi(0)$. $|(\delta,\varphi)| = |\varphi(0)| \leq \sup_{x\in K}|\varphi(x)|$, so it is a distribution of degree 0.

The δ measure is also a tempered distribution. For $g\in \mathcal S$, $|(\delta,g)| = |g(0)|\leq \sup_{x\in \mathbb{R}}|g(x)|$. This is very important in applications.

Operations on tempered distributions¶

We have studied the Fourier transform works well for Schwartz functions. In particular, we have the shifting theorem, derivative theorem, convolution theorem and inversion theorem, and the Poisson summation formula. We are going to see that these also generalizes to tempered distributions. But first, we need to make sense the operations.

Derivative¶

Let $T\in \mathcal D'$. The derivative of $T$ is defined by the distribution $T'$ such that $(T',\varphi) = -(T,\varphi')$ for any $\varphi \in \mathcal D$. One checks that $T'$ is also a distribution as follows: for every $\varphi \in \mathscr{C}_K^{\infty}(\mathbb{R})$, $|(T',\varphi)| = |(T,\varphi')| \leq C\sum_{0\leq k \leq N}\|\varphi'\|_{k} = C\sum_{0\leq k \leq N}\|\varphi\|_{k+1} \leq C\sum_{0\leq k\leq N+1}\|\varphi\|_k$.

As a motivation for the derivative, let’s consider the case when $T$ is determined a smooth function $f$, in this case, the derivative of $T_f$ should be $T_{f'}$. Then $(T_{f'},\varphi) = \int f'\varphi = -\int f\varphi' = -(T_f,\varphi')$.

We’ll frequently use the integration by parts formula in the special case when one of the function is compact support.

This allow us to define any derivatives of $f\in L^1_{loc}(\mathbb{R})$ in sense of distribution. Let $f\in L^1_{loc}(\mathbb{R})$, define its $k$-th order derivative by $(T_f^{(k)},\varphi) = (-1)^k (T,\varphi)$.

Let $T$ be a tempered distribution, then since $\mathcal S'\subset \mathcal D'$, $T$ is a distribution. Then $T$ has derivative $T'$ as a distribution. It’s straightforward to verify that $T'$ is also a tempered distribution. What is $(T',g)$ for $g\in \mathcal S$? We already know if $g$ has compact support then $(T',g) = -(T,g)$. But by the density lemma, $T$ is completely determined by its action on $\mathcal D$.

Multiplication by smooth function¶

Let $f$ be a smooth function, $T\in \mathcal D'$. We can define a new distribution $fT$. Again, to find it, we use our “try a function first” principle. If $T = T_g$, then $(fT_g,h) = \int fgh = (T_g,fh)$. So we can define $fT$ by its action $(fT,\varphi) = (T,f\varphi)$ for $\varphi \in \mathcal D$. To check that $fT$ is a distribution, for $\varphi \in \mathscr{C}_K^{\infty}(\mathbb{R})$, $|(fT,\varphi)| = |(T,f\varphi)| \leq C \sum_{0\leq k\leq N}\|f\varphi\|_{k,K} \leq C'\max_{0\leq k\leq N}{\|f\|_{k,K}}\sum_{0\leq k\leq N}\|\varphi\|_{k,K}$. Note that we used the Leibnitz rule to calculate $\|f\varphi\|_{k,K}$.

Similarly, if $T\in S'$, we can show that $fT\in S'$ for $f\in \mathscr{C}^{\infty}(\mathbb{R})$ with moderate growth condition (i.e. $f$ growth like polynomial). Then the density lemma implies

In particular, we have the multiplication $T \mapsto$ $2\pi i x T$ defined.

The Fourier transform of tempered distribution¶

For $T\in \mathcal S$, we’ll define its Fourier transform $\mathcal F T$. It should be another tempered distribution, and it should be defined by its action on $\mathcal D$. Then what should be $(\mathcal FT,\varphi)$ for $\varphi \in \mathcal D$? Still, let’s determine this from our “try a function” principle.

Let $f,g\in \mathcal S$. Then $\mathcal FT_{f}$ should be $T_{\mathcal Ff}$ because tempered distribution is generalization of functions. So we need to determine $(\mathcal FT_f,g) = (\mathcal Ff,g) = (T_f,?)$.

Lemma (multiplication identity)¶

For $f,g\in \mathcal S$, $\int \mathcal Ff \cdot g = \int f\cdot \mathcal Fg$.

Proof. This is just an application of Fubini theorem.

From the lemma we see that the $?$ above should be $\mathcal Fg$, and $(\mathcal FT_{f},g) = (T,\mathcal Fg)$.

We have not yet verified that $\mathcal FT$ is a tempered distribution. This is justified by the following lemma:

Lemma¶

For $f\in \mathcal S$, $\|\mathcal F f\|_N\leq C_N \cdot \|f\|_{N+2}$, where $\|f\|_{N}$ is given by $\sum_{0\leq k,l\leq N}\|f\|_{k,l}$.

Proof. See problem set 6.