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The Hermite-Gauss functions ($t\mapsto H_m(t)e^{-t^2/2}$) are known to be an orthonormal basis for $L^2(\mathbb{R})$, a fortiori linearly dense in $L^2(\mathbb{R})$, and all are in the Schwartz space (and hence in $L^p(\mathbb{R})$). These functions play an extremely important role in $L^2$ theory, Fourier transform theory, quantum mechanics, and elsewhere. I've never seen it mentioned whether or not the Hermite-Gauss functions are linearly dense in $L^1(\mathbb{R})$ - and, more generally, $L^p(\mathbb{R})$. It seems like a reasonable question to ask but Googling has not led me to an answer one way or another. My guess is that they are, in fact, not linearly dense in $L^p(\mathbb{R})$ except for $p=2$, but I don't have any clue as to how to prove that. Can anyone point me to this result or maybe give a clue as to why it is or isn't true?

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The functions $g_m(t) = H_m(t)\,e^{-t^2/2} $ just give an orthogonal base of $L^2(\mathbb{R})$, since: $$ \int_{-\infty}^{+\infty} f_m(t)^2\,dt = 2^m m! \sqrt{\pi}. $$ An orthonormal base is given by: $$ f_m(x) = \frac{1}{\sqrt{2^n n! \sqrt{\pi}}}\,H_m(x)\, e^{-x^2/2}. $$ We may notice that if $m$ is odd then $\int_\mathbb{R}f_m(x)\,dx = 0$, while: $$ \int_{\mathbb{R}} f_{2n}(x)\,dx = \frac{(2n)!}{n!}\sqrt{2\pi}\frac{1}{\sqrt{4^n (2n)! \sqrt{\pi}}}=\sqrt{\frac{2\sqrt{\pi}}{4^n}\binom{2n}{n}}\approx \sqrt{2}\cdot n^{-1/4}.$$ Moreover, we have $\left|\,f_m(x)\,\right|\leq\frac{1}{2}$ for every $m$ and every $x$, and $f_m$ is essentially zero outside $\left[-\frac{\pi}{2}\sqrt{m},\frac{\pi}{2}\sqrt{m}\right]$. So we know the behaviour of our base with respect to $L^1,L^2,L^\infty$. By interpolation or other techniques, it is not difficult to check that linear combinations of $f_n$s cannot be dense in $L^1$.

For instance, we may consider the function $g(x)=\frac{1}{\sqrt{x}}\cdot\mathbb{1}_{(0,1)}(x)$ that belongs to $L^1\setminus L^2$.

Every $f_n$ has a rather large support, so if we compute: $$ a_n = \int_{0}^{1} g(x)\,f_n(x)\,dx $$ we have that: $$ g_N(x)=\sum_{n=0}^{N} a_n\,f_n(x) $$ is a not-so-bad approximation of $g(x)$ over $[0,1]$, but it has a long tail, such that $g_N(x)$ does not converge to $g(x)$ in $L^1(\mathbb{R})$. To make this argument visually appealing, this is the situation for $N=10$:

$\hspace0.5in$enter image description here

This non-density argument is even more convincing (at least to me) if we notice that every $g_N$ is an entire function (of order 2) while $g$ is a compact-supported function with a branch-like singularity in a right neighbourhood of the origin. So the fact that $L^2$ and the Schwarz space are mapped into theirselves by the Fourier transform is really crucial for proving the density of the span of the "Hermite base".

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    $\begingroup$ I totally forgot to include the normalization constants. Thanks for the correction. I've seen interpolation but never thought it would apply to this situation. Thanks, Jack! $\endgroup$ – Cameron Williams Jul 19 '15 at 21:46
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    $\begingroup$ Re your last point: you could say the same about $L^2$. Those functions can decay even slower than those in $L^1$. $\endgroup$ – Cameron Williams Jul 19 '15 at 22:51
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    $\begingroup$ @CameronWilliams: you are right, so I deleted that lines. However, consider the following situation: take $\frac{1}{\sqrt{x}}$ supported on $[0,1]$. This function belongs to $L^1\setminus L^2$. Maybe a combination of $f_n$s can approximate $\frac{1}{\sqrt{x}}$ even uniformly over any compact subset of $(0,1)$, but this combination has a support way bigger than $[0,1]$, so it is not reasonable that this combination approximates $\frac{1}{\sqrt{x}}\cdot\mathbb{1}_{(0,1)}$ in $L^1(\mathbb{R})$. $\endgroup$ – Jack D'Aurizio Jul 19 '15 at 22:56
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    $\begingroup$ @CameronWilliams: one just need to compute the coefficients $a_n = \int_{0}^{1}\frac{f_n(x)}{\sqrt{x}}\,dx$ then plot $\sum_{n=0}^{N}a_n f_n(x)$ over $[0,2]$ to see that we have a slow convergence on $[0,1]$ and massive tails on $[1,2]$. I wonder if it is the case to include this example in my answer. $\endgroup$ – Jack D'Aurizio Jul 19 '15 at 23:09
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    $\begingroup$ @JackD'Aurizio I'm not the downvoter (and I'm a couple years late), but it seems that your answer only shows that the Hermite polynomials are not a Schauder basis: they could still have a dense span (as is the case in the Stone-Weierstrauss theorem). $\endgroup$ – Strants Aug 17 '18 at 21:59
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The ordinary heat equation is $$ \frac{\partial F}{\partial t}=\frac{\partial^{2}F}{\partial x^{2}},\\ F(0,x)=f(x). $$ The initial heat distribution is $f$. The ordinary heat kernel is the Guassian, and the resulting time evolution operator $T(t)=e^{tL}$ is a constractive $C_0$ semigroup on every $L^{p}(\mathbb{R})$ for $1 \le p < \infty$; the fact that it is $C_{0}$ gives $$ \|e^{tL}f-f\|_{p}\rightarrow 0 \mbox{ as } t\downarrow 0,\;\; 1 \le p < \infty. $$ That gives a nice approximation. In fact, this approximation technique goes back to Weierstrass in his original proof of the Weierstrass Approximation Theorem.

The Hermite functions $h_{n}(x)=H_{n}(x)e^{-x^{2}/2}$ are the $L^{2}$ eigenfunctions of $$ Lf = -\frac{d^{2}f}{dx^{2}}+x^{2}f $$ with eigenvalues $\lambda = 2n+1$ for $n=0,1,2,3,\cdots$. The Hermite functions $\{ h_{n} \}_{n=0}^{\infty}$ form a complete orthonormal basis of $L^{2}(\mathbb{R})$. The heat equation associated with $L$ is $$ \frac{\partial F}{\partial t}=\frac{\partial^{2}F}{\partial^{2}x}-x^{2}F,\\ F(0,x)=f(x). $$ This heat equation is better behaved in many ways than the ordinary heat equation because $-x^{2}F$ pulls heat out of the system near $\pm\infty$ for positive $F$. The time evolution solution operator $T(t)=e^{tL}$ in this case is $$ T(t)f = \sum_{n=0}^{\infty}e^{-(2n+1)t}(f,h_n)h_n. $$ So the approximation problem by Hermite functions is closely related to the continuity properties of the heat solution at $t=0$. That's why one studies the Hermite kernel function $$ K(r,x,y)=\sum_{n=0}^{\infty}r^{n}h_{n}(x)h_{n}(y), $$ which has an explicit representation as a bivariate Gaussian: $$ K(r,x,y) = \frac{1}{\sqrt{\pi(1-r^{2})}} \exp\left\{-\frac{1}{4}\frac{1-r}{1+r}(x+y)^{2}-\frac{1}{4}\frac{1+r}{1-r}(x-y)^{2}\right\}. $$ So the approximation problem can be studied by looking at the question $$ f\;\; ? = ?\;\; \lim_{r\downarrow 0}\int_{-\infty}^{\infty}K(r,x,y)f(y)dy = \lim_{r\downarrow 0}\sum_{n=0}^{\infty}r^{n}(f,h_n)h_n(x). $$

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  • $\begingroup$ This is really interesting, never would have thought to appeal to the heat equation, but I have one qualm. In your last line, are you suggesting that they are in fact dense in $L^p$? Also, are you missing a factor of $r$ (or something) in your expression for $K$ to account for the $-t$ part in the exponent? $\endgroup$ – Cameron Williams Jul 19 '15 at 22:30
  • $\begingroup$ @CameronWilliams : I left that open, but was guessing that the evolution equation probably did lead to a $C_0$ semigroup. I'm surprised that it wouldn't. Usually such physical problems are very well-posed. You must get pointwise convergence as $t\downarrow 0$ for any smooth $f$. $\endgroup$ – DisintegratingByParts Jul 19 '15 at 22:37
  • $\begingroup$ Hmm. Interesting. I guess that means that yours and Jack's answers are conflicting. I'll have to think about this some more. $\endgroup$ – Cameron Williams Jul 19 '15 at 22:38
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    $\begingroup$ @CameronWilliams : I don't think that conclusion would follow. Superficially speaking, there's a conspicuous lack of $L^p$ information out there. So I wouldn't be hasty to conclude anything. We could guess that the basis is not going to be a Schauder basis of $L^{1}$. $\endgroup$ – DisintegratingByParts Jul 19 '15 at 22:50
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    $\begingroup$ Ahh so it takes $L^p$ data and turns it into an $L^2$ function? That's still pretty neat. Don't delete your answer. It's quite beautiful. $\endgroup$ – Cameron Williams Jul 19 '15 at 23:36
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Hermite functions are dense in $L^1[\mathbb{R}]$, but we cannot use coefficients computed as above. Every $L^1[\mathbb{R}]$ function can be approximated in $L^1[\mathbb{R}]$ by continuous bounded functions of compact support - so they are also in $L^2[\mathbb{R}]$ and can be approximated by linear combinations of Hermite functions.

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  • $\begingroup$ Welcome on the MathSE! Note, this site supports Latex, just write $L^1(\mathbb{R})$ and you get $L^1(\mathbb{R})$. $\endgroup$ – peterh - Reinstate Monica Aug 17 '18 at 21:23

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