The Fourier Transform of exp(-cx^2) and Its Convolution

We develop two almost straightforward way to compute the Fourier transform of $\exp(-cx^2)$, in the sense that any contour integration and the calculus of residues are not required at all. The first cool approach enables us to think about these elementary concepts much deeper, so I highly recommend to study this approach as long as you are familiar with ODE of first order.
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Basic Facts of Semicontinuous Functions


We are restricting ourselves into $\mathbb{R}$ endowed with normal topology. Recall that a function is continuous if and only if for any open set $U \subset \mathbb{R}$, we have

to be open. One can rewrite this statement using $\varepsilon-\delta$ language. To say a function $f: \mathbb{R} \to \mathbb{R}$ continuous at $f(x)$, we mean for any $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$f$ is continuous on $\mathbb{R}$ if and only if $f$ is continuous at every point of $\mathbb{R}$.

If $(x-\delta,x+\delta)$ is replaced with $(x-\delta,x)$ or $(x,x+\delta)$, we get left continuous and right continuous, one of which plays an important role in probability theory.

But the problem is, sometimes continuity is too strong for being a restriction, but the ‘direction’ associated with left/right continuous functions are unnecessary as well. For example the function

is neither left nor right continuous (globally), but it is a thing. Left/right continuity is not a perfectly weakened version of continuity. We need something different.

Definition of semicontinuous

Let $f$ be a real (or extended-real) function on $\mathbb{R}$. The semicontinuity of $f$ is defined as follows.


is open for all real $\alpha$, we say $f$ is lower semicontinuous.


is open for all real $\alpha$, we say $f$ is upper semicontinuous.

Is it possible to rewrite these definitions à la $\varepsilon-\delta$? The answer is yes if we restrict ourselves in metric space.

$f: \mathbb{R} \to \mathbb{R}$ is upper semicontinuous at $x$ if, for every $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$f: \mathbb{R} \to \mathbb{R}$ is lower semicontinuous at $x$ if, for every $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

Of course, $f$ is upper/lower semicontinuous on $\mathbb{R}$ if and only if it is so on every point of $\mathbb{R}$. One shall find no difference between the definitions in different styles.

Relation with continuous functions

Here is another way to see it. For the continuity of $f$, we are looking for arbitrary open subsets $V$ of $\mathbb{R}$, and $f^{-1}(V)$ is expected to be open. For the lower/upper semicontinuity of $f$, however, the open sets are restricted to be like $(\alpha,+\infty]$ and $[-\infty,\alpha)$. Since all open sets of $\mathbb{R}$ can be generated by the union or intersection of sets like $[-\infty,\alpha)$ and $(\beta,+\infty]$, we immediately get

$f$ is continuous if and only if $f$ is both upper semicontinuous and lower semicontinuous.

Proof. If $f$ is continuous, then for any $\alpha \in \mathbb{R}$, we see $[-\infty,\alpha)$ is open, and therefore

has to be open. The upper semicontinuity is proved. The lower semicontinuity of $f$ is proved in the same manner.

If $f$ is both upper and lower semicontinuous, we see

is open. Since every open subset of $\mathbb{R}$ can be written as a countable union of segments of the above types, we see for any open subset $V$ of $\mathbb{R}$, $f^{-1}(V)$ is open. (If you have trouble with this part, it is recommended to review the definition of topology.) $\square$


There are two important examples.

  1. If $E \subset \mathbb{R}$ is open, then $\chi_E$ is lower semicontinuous.
  2. If $F \subset \mathbb{R}$ is closed, then $\chi_F$ is upper semicontinuous.

We will prove the first one. The second one follows in the same manner of course. For $\alpha<0$, the set $A=\chi_E^{-1}((\alpha,+\infty])$ is equal to $\mathbb{R}$, which is open. For $\alpha \geq 1$, since $\chi_E \leq 1$, we see $A=\varnothing$. For $0 \leq \alpha < 1$ however, the set of $x$ where $\chi_E>\alpha$ has to be $E$, which is still open.

When checking the semicontinuity of a function, we check from bottom to top or top to bottom. The function $\chi_E$ is defined by

Addition of semicontinuous functions

If $f_1$ and $f_2$ are upper/lower semicontinuous, then so is $f_1+f_2$.

Proof. We are going to prove this using different tools. Suppose now both $f_1$ and $f_2$ are upper semicontinuous. For $\varepsilon>0$, there exists some $\delta_1>0$ and $\delta_2>0$ such that

Proof. If we pick $\delta=\min(\delta_1,\delta_2)$, then we see for all $t \in (x-\delta,x+\delta)$, we have

The upper semicontinuity of $f_1+f_2$ is proved by considering all $x \in \mathbb{R}$.

Now suppose both $f_1$ and $f_2$ are lower semicontinuous. We have an identity by

The set on the right side is always open. Hence $f_1+f_2$ is lower semicontinuous. $\square$

However, when there are infinite many semicontinuous functions, things are different.

Let $\{f_n\}$ be a sequence of nonnegative functions on $\mathbb{R}$, then

  • If each $f_n$ is lower semicontinuous, then so is $\sum_{1}^{\infty}f_n$.
  • If each $f_n$ is upper semicontinuous, then $\sum_{1}^{\infty}f_n$ is not necessarily upper semicontinuous.

Proof. To prove this we are still using the properties of open sets. Put $g_n=\sum_{1}^{n}f_k$. Now suppose all $f_k$ are lower. Since $g_n$ is a finite sum of lower functions, we see each $g_n$ is lower. Let $f=\sum_{n}f_n$. As $f_k$ are non-negative, we see $f(x)>\alpha$ if and only if there exists some $n_0$ such that $g_{n_0}(x)>\alpha$. Therefore

The set on the right hand is open already.

For the upper semicontinuity, it suffices to give a counterexample, but before that, we shall give the motivation.

As said, the characteristic function of a closed set is upper semicontinuous. Suppose $\{E_n\}$ is a sequence of almost disjoint closed set, then $E=\cup_{n\geq 1}E_n$ is not necessarily closed, therefore $\chi_E=\sum\chi_{E_n}$ (a.e.) is not necessarily upper semicontinuous. Now we give a concrete example. Put $f_0=\chi_{[1,+\infty]}$ and $f_n=\chi_{E_n}$ for $n \geq 1$ where

For $x > 0$, we have $f=\sum_nf_n \geq 1$. Meanwhile, $f^{-1}([-\infty,1))=[-\infty,0]$, which is not open. $\square$

Notice that $f$ can be defined on any topological space here.

Maximum and minimum

There is one fact we already know about continuous functions.

If $X$ is compact, $f: X \to \mathbb{R}$ is continuous, then there exists some $a,b \in X$ such that $f(a)=\min f(X)$, $f(b)=\max f(X)$.

In fact, $f(X)$ is compact still. But for semicontinuous functions, things will be different but reasonable. For upper semicontinuous functions, we have the following fact.

If $X$ is compact and $f: X \to (-\infty,+\infty)$ is upper semicontinuous, then there exists some $a \in X$ such that $f(a)=\max f(X)$.

Notice that $X$ is not assumed to hold any other topological property. It can be Hausdorff or Lindelöf, but we are not asking for restrictions like this. The only property we will be using is that every open cover of $X$ has a finite subcover. Of course, one can replace $X$ with any compact subset of $\mathbb{R}$, for example, $[a,b]$.

Proof. Put $\alpha=\sup f(X)$, and define

If $f$ attains no maximum, then for any $x \in X$, there exists some $n \geq 1$ such that $f(x)<\alpha-\frac{1}{n}$. That is, $x \in E_n$ for some $n$. Therefore $\bigcup_{n \geq 1}E_n$ covers $X$. But this cover has no finite subcover of $X$. A contradiction since $X$ is compact. $\square$

Approximating integrable functions

This is a comprehensive application of several properties of semicontinuity.

(Vitali–Carathéodory theorem) Suppose $f \in L^1(\mathbb{R})$, where $f$ is a real-valued function. For $\varepsilon>0$, there exist some functions $u$ and $v$ on $\mathbb{R}$ such that $u \leq f \leq v$, $u$ is an upper semicontinuous function bounded above, and $v$ is lower semicontinuous bounded below, and

It suffices to prove this theorem for $f \geq 0$ (of course $f$ is not identically equal to $0$ since this case is trivial). Since $f$ is the pointwise limit of an increasing sequence of simple functions $s_n$, can to write $f$ as

By putting $t_1=s_1$, $t_n=s_n-s_{n-1}$ for $n \geq 2$, we get $f=\sum_n t_n$. We can write $f$ as

where $E_k$ is measurable for all $k$. Also, we have

and the series on the right hand converges (since $f \in L^1$. By the properties of Lebesgue measure, there exists a compact set $F_k$ and an open set $V_k$ such that $F_k \subset E_k \subset V_k$ and $c_km(V_k-F_k)<\frac{\varepsilon}{2^{k+1}}$. Put

(now you can see $v$ is lower semicontinuous and $u$ is upper semicontinuous). The $N$ is chosen in such a way that

Since $V_k \supset E_k$, we have $\chi_{V_k} \geq \chi_{E_k}$. Therefore $v \geq f$. Similarly, $f \geq u$. Now we need to check the desired integral inequality. A simple recombination shows that

If we integrate the function above, we get

This proved the case when $f \geq 0$. In the general case, we write $f=f^{+}-f^{-}$. Attach the semicontinuous functions to $f^{+}$ and $f^{-}$ respectively by $u_1 \leq f^{+} \leq v_1$ and $u_2 \leq f^{-} \leq v_2$. Put $u=u_1-v_2$, $v=v_1-u_2$. As we can see, $u$ is upper semicontinuous and $v$ is lower semicontinuous. Also, $u \leq f \leq v$ with the desired property since

and the theorem follows. $\square$


Indeed, the only property about measure used is the existence of $F_k$ and $V_k$. The domain $\mathbb{R}$ here can be replaced with $\mathbb{R}^k$ for $1 \leq k < \infty$, and $m$ be replaced with the respective $m_k$. Much more generally, the domain can be replaced by any locally compact Hausdorff space $X$ and the measure by any measure associated with the Riesz-Markov-Kakutani representation theorem on $C_c(X)$.

Is the reverse approximation always possible?

The answer is no. Consider the fat Cantor set $K$, which has Lebesgue measure $\frac{1}{2}$. We shall show that $\chi_K$ can not be approximated below by a lower semicontinuous function.

If $v$ is a lower semicontinuous function such that $v \leq \chi_K$, then $v \leq 0$.

Proof. Consider the set $V=v^{-1}((0,1])=v^{-1}((0,+\infty))$. Since $v \leq \chi_K$, we have $V \subset K$. We will show that $V$ has to be empty.

Pick $t \in V$. Since $V$ is open, there exists some neighbourhood $U$ containing $t$ such that $U \subset V$. But $U=\varnothing$ since $U \subset K$ and $K$ has an empty interior. Therefore $V = \varnothing$. That is, $v \leq 0$ for all $x$. $\square$

Suppose $u$ is an upper semicontinuous function such that $u \geq f$. For $\varepsilon=\frac{1}{2}$, we have

This example shows that there exist some integrable functions that are not able to reversely approximated in the sense of the Vitali–Carathéodory theorem.



洛必达法则相比甚至不少高中生甚至初中生都听说过,知道怎么进行简单的应用。简单点说,处理$\frac{0}{0}$的函数时,对上下进行求导,可能会很大程度上简化计算。但是洛必达法则为什么能奏效? 能不能用严格的数理语言进行论证? 这是这篇文章需要解决的。



设$f$和$g$为$(a,b)$上的实可微函数,且在$(a,b)$区间上总有$g’(x) \ne 0$. 设


其中$x \to a$自然也可以换成$x \to b$. 这里把发散到无穷也看作是极限,也就是说$-\infty \le A \le \infty$且$-\infty \le a < b \le +\infty$.



洛必达法则首次出现于1696年洛必达的 Analyse des Infiniment Petits pour l’Intelligence des Lignes Courbes 一书中。这本书当然以”洛必达法则”闻名于世。证明是这样完成的:

这个证明很好理解,线性近似展开,再考虑到$f(a)=g(a)=0$就得到结果。但是这个做法肯定是不合适的,$dx$在这里非常模糊,也不方便表达$x\to\infty$的情况。关于历史内容可以参见 The Historical Development of the Calculus 一书。










这个证明中,我们会利用柯西中值定理(GMVT)对所有的情况进行完整的证明,这期间涉及到一些不等式运算技巧。证明来自W. Rudin的 Principles Of Mathematical Analysis,我会在其中加上一些额外的解释。关于$g(x) \to \pm\infty$的情况,我们在此只讨论$+\infty$.

情况1: $-\infty\leq{A}<+\infty$

选取实数$q>A$,再选取$\varepsilon>0$使得$A+\varepsilon<q$. 因为$\frac{f’(x)}{g’(x)}\to{A}$,根据极限的定义,必定有实数$\delta\in(0,b-a)$,使得对于所有$a<x<a+\delta$,始终有$-\varepsilon<\frac{f’(x)}{g’(x)}-A<\varepsilon$. 特别地,


最后一个不等式成立是因为$t\in(x,y)\subset(a,a+\delta)$,而在$(a,a+\delta)$中这个不等式成立。接下来,我们根据$g(x)$在$x \to a$时的取值,分别讨论,会发现结果其实类似.

情况1.1: $g(x)\to 0$且$f(x) \to 0$

在不等式$\eqref{(A)}$中,令$x \to a$,会发现有:

更正式地说,对任意的$q>A$和$0< \varepsilon< q-A$,都存在$\delta > 0$,使得对任意的$a<y<a+\delta$,均满足不等式


情况1.2: $g(x)\to+\infty$

固定不等式(A)中的$ y $. 因为$ g(x) \to +\infty $,在$ (a,y) $一定存在$ c_1 $使得对于任意$ a< x < c_1 $,总是有$ g(x) > g(y) $和$ g(x) > 0 $同时成立。将不等式(A)两边同时乘以$ [g(x)-g(y)]/g(x) $,我们得到


根据$g(x) \to +\infty$的定义,存在$c_2 \in (a,c_1)$使得:


不等式$\eqref{(B)}$和$\eqref{(D)}$都只说明,存在$c\in(a,b)$使得对于所有$x\in(a,c)$,满足$\frac{f(x)}{g(x)}<q$.但是$\frac{f(x)}{g(x)}$与$A$的关系我们并不知道。但是,如果$A=-\infty$,那么我们已经证明了$\frac{f(x)}{g(x)} \to -\infty$的成立。

情况2: $-\infty<{A}\leq+\infty$

这个情况是和情况1完全类似的。同理可证,对任意$p$,当且仅当$p<A$时,总有$c’\in(a,b)$,使得对于所有$x\in(a,c’)$,满足$p<\frac{f(x)}{g(x)}$. 如果$A=+\infty$,那么我们已经证明了$\frac{f(x)}{g(x)} \to +\infty$的情况。

除去$A=\pm\infty$,综合情况1和2,我们发现,对任意的$p,q$满足$p<A<q$,若取$c_0=\min\{c,c’\}$,则对于任意$x \in (a,c_0)$,一定有

这其实就等价于$\lim_{x \to a}\frac{f(x)}{g(x)}=A$. $\square$



假设它无意义. 如果有$g(x)=g(y)$,那么有$a<{x}<t<y<b$使得$g’(t)=0$, 而这与原假设矛盾.