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The Fourier transform of an $L^1$ function is defined by $$\hat{f}(y)=\int_\mathbb{R}f(x)e^{-ixy}dx$$

Is it true that for functions $f,g\in L^1$, if $\hat{f}=\hat{g}$, then $f=g$?

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    $\begingroup$ Yes they are equivalent, almost everywhere. $\endgroup$ Nov 17, 2013 at 3:30
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    $\begingroup$ @CameronWilliams Okay, how to show that? $\endgroup$
    – Kunal
    Nov 17, 2013 at 3:34

3 Answers 3

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$$ \int_{\mathbb R} e^{-a|x|^2/2 + i x x_0} \hat f(x) \, dx = \int_{\mathbb R} f(y) \int_{\mathbb R} \frac1{2\pi} e^{-a|x|^2/2} e^{-ix(y-x_0)} \, dx \, dy = \int_{\mathbb R} \frac1{\sqrt{2 \pi a}} e^{-|y-x_0|^2/2a} f(y) \, dy .$$ The first equality is by substitution and Fubini. The second uses standard tables of Fourier transforms. The last quantity converges to $f(x_0)$ in $L_1$ as $a\to 0^+$ (see first that it converges if $f$ is a continuous and compactly supported, then use the usual tricks of approximating $f$ by a such a function). I might be off by a factor of $2\pi$.

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  • $\begingroup$ By "converges" I suppose you mean converges as $a\to0^+$. $\endgroup$
    – Hans
    Oct 17, 2017 at 23:25
  • $\begingroup$ @Hans yes, that's right. $\endgroup$ Oct 18, 2017 at 0:00
  • $\begingroup$ It would be better to specify that for clarification, even though it is trivial. $\endgroup$
    – Hans
    Oct 18, 2017 at 0:05
  • $\begingroup$ Actually, the exponent of the right-hand side is not correct. It should be $-\frac12\big[(\frac y{\sqrt{a}}+i\sqrt a x_0)^2+ax_0^2\big]$, which only makes RHS $\to f(0)$ as $a\to 0^+$. The argument needs to be modified, because the shift $f(\cdot+x_0)$ results in multiplication of $\hat f(x)$ by $e^{ixx_0}$ and $f$ and $\hat f$ cannot both shift. You can either make the argument for $x_0=0$ then apply the shift operator $T_{x_0}$ to $f$, i.e., $T_{x_0}f(\cdot)=f(\cdot+x_0)$, or let the LHS= $\displaystyle\int_{\mathbb R} e^{-ax^2/2+ixx_0} \hat f(x) \, dx$. $\endgroup$
    – Hans
    Oct 19, 2017 at 2:56
  • $\begingroup$ @Hans Thank you for the correction. I fixed it. $\endgroup$ Oct 19, 2017 at 13:34
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EDIT: This only works for $f,g\in L^1\cap L^2$

HINT: Use the inverse fourier transform: $$f(x)=\frac{1}{2\pi}\int_\mathbb{R}\hat{f}(y)e^{ixy}dy$$

and let $\hat{h}=\hat{f}-\hat{g}\equiv0$, then conclude.

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    $\begingroup$ This doesn't quite work. The inverse Fourier transform is not necessarily defined if $f,g\in L^1$. $\endgroup$ Nov 17, 2013 at 3:32
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    $\begingroup$ @CameronWilliams: Thanks for your reminder, I have added an assumpton. $\endgroup$ Nov 17, 2013 at 3:38
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As an extension to Stephen's answer. Here's the justification that if $f\in L^{\infty}(\Bbb{R}^{n})\cap C(\Bbb{R}^{n})$ then

$$\lim_{a\to 0^{+}}\frac{1}{(\pi a)^{n/2}}\int_{\Bbb{R}^{n}}f(x)e^{-|x-x_{0}|^{2}/a}\,dx=f(x_{0})$$

WLOG we will assume $x_{0}=0$ as the proof is similar in the general case.

Now, note that $\displaystyle \frac{1}{(\pi a)^{n/2}}\int e^{-|x|^{2}/a}\,dx=1$

Hence we should be looking at, $$\bigg|\frac{1}{(\pi a)^{n/2}}\int_{\Bbb{R}^{n}}f(x)e^{-|x|^{2}/a}dx-\frac{1}{(\pi a)^{n/2}}\int_{\Bbb{R}^{n}}f(0)e^{-|x|^{2}/a}dx\bigg|$$

Let $\epsilon>0$ be arbitrary. Choose an $r$ such that $|f(x)-f(0)|<\epsilon$ for all $x\in B(0,r)$

The integral then can be written as,

$$\frac{1}{(\pi a)^{n/2}}\int_{\Bbb{R}^{n}\setminus B(0,r)}(f(x)-f(0))e^{-|x|^{2}/a}dx+\frac{1}{(\pi a)^{n/2}}\int_{B(0,r)}(f(x)-f(0))e^{-|x|^{2}/a}dx$$

Now as $f\in L^{\infty}$ we have that $|\frac{1}{(\pi a)^{n/2}}(f(x)-f(0))e^{-|x|^{2}/a}\cdot\mathbf{1}_{\Bbb{R}^{n}\setminus B(0,r)}|\leq C \frac{1}{(\pi r)^{n/2}}e^{-|x|^{2}}$ for all $a\leq 1$. The RHS is integrable in $\Bbb{R}^{n}$.

But $\frac{1}{(\pi a)^{n/2}}(f(x)-f(0))e^{-|x|^{2}/a}\xrightarrow{a\to 0^{+}} 0$ for all $x\in \Bbb{R}^{n}\setminus B(0,r)$. Thus by DCT, the first integral is less than $\epsilon$ for all $a<\delta(r)$ where $\delta(r,\epsilon)$ is some number depending on $r$ and $\epsilon$ and hence only on $\epsilon$.

Also we have that \begin{align}\bigg|\frac{1}{(\pi a)^{n/2}}\int_{B(0,r)}(f(x)-f(0))e^{-|x|^{2}/a}dx\bigg|&\leq \int_{B(0,r)}\bigg|\frac{1}{(\pi a)^{n/2}}\int_{B(0,r)}(f(x)-f(0))e^{-|x|^{2}/a}\bigg|dx\\\\ &\leq \int_{B(0,r)}\frac{1}{(\pi a)^{n/2}}\int_{B(0,r)}\bigg|f(x)-f(0)\bigg|e^{-|x|^{2}/a}dx\\\\ &\leq \frac{1}{(\pi a)^{n/2}}\int_{B(0,r)}\epsilon \cdot e^{-|x|^{2}/a}\,dx\\\\ &\leq \frac{\epsilon}{(\pi a)^{n/2}}\int_{\Bbb{R}^{n}}e^{-|x|^{2}/a}\,dx=\epsilon \end{align}

Thus we have that $$\lim_{a\to 0^{+}}\frac{1}{(\pi a)^{n/2}}\int_{\Bbb{R}^{n}}f(x)e^{-|x|^{2}/a}\,dx=f(0)$$

Now, the general case has the same proof. We just consider the ball $B(x_{0},r)$ instead and repeat the steps.

Now that this holds for all $f\in L^{\infty}(\Bbb{R}^{n})\cap C(\Bbb{R}^{n})$, we can by exploiting the density of such functions conclude the same result for $f\in L^{1}(\Bbb{R}^{n})$

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