Copy edited. Fixed the question formation - missing auxiliary verb - see e.g. <https://www.youtube.com/watch?v=t4yWEt0OSpg&t=1m49s> (see also <https://www.youtube.com/watch?v=kS5NfSzXfrI> (QUASM)). Removed meta information (this is implied by posting here).

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edit approved Oct 24, 2023 at 4:19

Peter Mortensen

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In what sense does the Hermite expansion of a bounded smooth function convergesconverge?

Let $f : \mathbb{R} \to \mathbb{C}$ be a smooth and bounded function.

If we denote by $\{ H_n(x) \}$ the sequence of normalized Hermite polynomials, then the Hermite expansion of $f$ is defined as

\begin{equation} \sum_{n=0}^\infty d_n H_n(x) \end{equation} where \begin{equation} d_n=\frac{1}{\sqrt{2\pi}n!}\int_{\mathbb{R}}f(x)H_n(x)e^{-x^2/2}dx \end{equation}

According to the following link : Convergence of orthogonal polynomial expansions

Convergence of orthogonal polynomial expansions, the Hermite expansion converges uniformly to $f$ if $\lVert Hf \rVert_{L^2} < \infty$.

However, if I just assume that $f$ is smooth and bounded, can I still say something about convergence of its Hermite expansion?

I tried to look for references myself, but cannotI could not find anything relevant.

Could anyone please help me?

In what sense the Hermite expansion of a bounded smooth function converges?

Let $f : \mathbb{R} \to \mathbb{C}$ be a smooth and bounded function.

If we denote by $\{ H_n(x) \}$ the sequence of normalized Hermite polynomials, then the Hermite expansion of $f$ is defined as \begin{equation} \sum_{n=0}^\infty d_n H_n(x) \end{equation} where \begin{equation} d_n=\frac{1}{\sqrt{2\pi}n!}\int_{\mathbb{R}}f(x)H_n(x)e^{-x^2/2}dx \end{equation}

According to the following link : Convergence of orthogonal polynomial expansions

the Hermite expansion converges uniformly to $f$ if $\lVert Hf \rVert_{L^2} < \infty$.

However, if I just assume that $f$ is smooth and bounded, can I still say something about convergence of its Hermite expansion?

I tried to look for references myself, but cannot find anything relevant.

Could anyone please help me?

In what sense does the Hermite expansion of a bounded smooth function converge?

Let $f : \mathbb{R} \to \mathbb{C}$ be a smooth and bounded function.

If we denote by $\{ H_n(x) \}$ the sequence of normalized Hermite polynomials, then the Hermite expansion of $f$ is defined as

\begin{equation} \sum_{n=0}^\infty d_n H_n(x) \end{equation} where \begin{equation} d_n=\frac{1}{\sqrt{2\pi}n!}\int_{\mathbb{R}}f(x)H_n(x)e^{-x^2/2}dx \end{equation}

According to Convergence of orthogonal polynomial expansions, the Hermite expansion converges uniformly to $f$ if $\lVert Hf \rVert_{L^2} < \infty$.

However, if I just assume that $f$ is smooth and bounded, can I still say something about convergence of its Hermite expansion?

I tried to look for references myself, but I could not find anything relevant.

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occurred Oct 23, 2023 at 21:00

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edited Oct 23, 2023 at 12:59

Isaac

3.5k
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22

Let $f : \mathbb{R} \to \mathbb{C}$ be a smooth and bounded function.

If we denote by $\{ H_n(x) \}$ the sequence of normalized Hermite polynomials, then the Hermite expansion of $f$ is defined as \begin{equation} \sum_{n=1}^\infty d_n H_n(x) \end{equation}\begin{equation} \sum_{n=0}^\infty d_n H_n(x) \end{equation} where \begin{equation} d_n=\frac{1}{\sqrt{2\pi}n!}\int_{\mathbb{R}}f(x)H_n(x)e^{-x^2/2}dx \end{equation}

According to the following link : Convergence of orthogonal polynomial expansions

the Hermite expansion converges uniformly to $f$ if $\lVert Hf \rVert_{L^2} < \infty$.

However, if I just assume that $f$ is smooth and bounded, can I still say something about convergence of its Hermite expansion?

I tried to look for references myself, but cannot find anything relevant.

Could anyone please help me?

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asked Oct 23, 2023 at 12:52

Isaac

3.5k
16
22

In what sense the Hermite expansion of a bounded smooth function converges?

Let $f : \mathbb{R} \to \mathbb{C}$ be a smooth and bounded function.

If we denote by $\{ H_n(x) \}$ the sequence of normalized Hermite polynomials, then the Hermite expansion of $f$ is defined as \begin{equation} \sum_{n=1}^\infty d_n H_n(x) \end{equation} where \begin{equation} d_n=\frac{1}{\sqrt{2\pi}n!}\int_{\mathbb{R}}f(x)H_n(x)e^{-x^2/2}dx \end{equation}

According to the following link : Convergence of orthogonal polynomial expansions

the Hermite expansion converges uniformly to $f$ if $\lVert Hf \rVert_{L^2} < \infty$.

However, if I just assume that $f$ is smooth and bounded, can I still say something about convergence of its Hermite expansion?

I tried to look for references myself, but cannot find anything relevant.

Could anyone please help me?

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In what sense does the Hermite expansion of a bounded smooth function convergesconverge?

In what sense the Hermite expansion of a bounded smooth function converges?

In what sense does the Hermite expansion of a bounded smooth function converge?

In what sense the Hermite expansion of a bounded smooth function converges?