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No, and a simple example goes as follows: $$ \begin{split} K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) & =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\ & \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N \end{split} $$ where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in this answer is not.

The standard kernel theorem ([1] chapter 1, §1.3 pp. 11-20 and §3.5 pp. 73-79) guarantees "only" that there is a distribution which does the job, but does not guarantee that there exists an integral representation, though this can be true for general $n$-linear functionals on a particular function space, or for particular functionals on general function spaces: also, the integral representation may not have the form \eqref{1}. For example, in the case of linear functionals, an integral representation as a contour integral $$ f(\varphi)=\oint_\gamma v(z)\varphi(z)\mathrm{d}z\quad \varphi\in \mathscr{H\!\!o\!l}(D) $$ where

• $v(z)$ is the Fantappiè indicatrix of the functional, i.e. the (in general meromorphic) function $$ v(z)=f\left(\frac{1}{\zeta-z}\right)\quad(\zeta\text{ is the "inner" variable}) $$
• $\gamma\subset D\subseteq \Bbb C$ is a curve enclosing all singularities of $v$.

holds for every linear analytic functional $f\in\mathscr{O}(\Bbb C)$. And there are also more general "integral" representations (if you allow for the kernel $K$ to be a Radon Measure) which include large classes of linear functionals and more general generalized functions: see for example the paper [2].

References

[1] Gel’fand, I. M.; Vilenkin, N. Ya., Generalized functions. Vol. 4: Applications of harmonic analysis, Translated from the Russian by Amiel Feinstein. (English) New York and London: Academic Press. XIV, 384 p. (1964), MR0173945, Zbl 0136.11201.

[2] Kaneko, Akira, "Representation of hyperfunctions by measures and some of its applications", (English) Journal of the Faculty of Science, Section I A 19, 321-352 (1972), MR0336328, Zbl 0247.35007.

No, and a simple example goes as follows: $$ \begin{split} K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) & =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\ & \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N \end{split} $$ where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in this answer is not.

The standard kernel theorem ([1] chapter 1, §1.3 pp. 11-20 and §3.5 pp. 73-79) guarantees "only" that there is a distribution which does the job, but does not guarantee that there exists an integral representation, though this can be true for general $n$-linear functionals on a particular function space, or for particular functionals on general function spaces: also, the integral representation may not have the form \eqref{1}. For example, in the case of linear functionals, an integral representation as a contour integral $$ f(\varphi)=\oint_\gamma v(z)\varphi(z)\mathrm{d}z\quad \varphi\in \mathscr{H\!\!o\!l}(D) $$ where

• $v(z)$ is the Fantappiè indicatrix of the functional, i.e. the (in general meromorphic) function $$ v(z)=f\left(\frac{1}{\zeta-z}\right)\quad(\zeta\text{ is the "inner" variable}) $$
• $\gamma\subset D\subseteq \Bbb C$ is a curve enclosing all singularities of $v$.

holds for every linear analytic functional $f\in\mathscr{O}^\prime\!(\Bbb C)$. And there are also more general "integral" representations (if you allow for the kernel $K$ to be a Radon Measure) which include large classes of linear functionals and more general generalized functions: for example, see the paper [2].

References

[1] Gel’fand, I. M.; Vilenkin, N. Ya., Generalized functions. Vol. 4: Applications of harmonic analysis, Translated from the Russian by Amiel Feinstein. (English) New York and London: Academic Press. XIV, 384 p. (1964), MR0173945, Zbl 0136.11201.

[2] Kaneko, Akira, "Representation of hyperfunctions by measures and some of its applications", (English) Journal of the Faculty of Science, Section I A 19, 321-352 (1972), MR0336328, Zbl 0247.35007.

No, and a simple example goes as follows: $$ \begin{split} K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) & =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\ & \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N \end{split} $$ where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in this answer is not.

The standard kernel theorem ([1] chapter 1, §1.3 pp. 11-20 and §3.5 pp. 73-79) guarantees "only" that there is a distribution which does the job, but does not guarantee that there exists an integral representation, though this can be true for general $n$-linear functionals on a particular function space, or for particular functionals on general function spaces: also, the integral representation may not have the form \eqref{1}.

References

[1] Gel’fand, I. M.; Vilenkin, N. Ya., Generalized functions. Vol. 4: Applications of harmonic analysis, Translated from the Russian by Amiel Feinstein. (English) New York and London: Academic Press. XIV, 384 p. (1964), MR0173945, Zbl 0136.11201.

No, and a simple example goes as follows: $$ \begin{split} K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) & =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\ & \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N \end{split} $$ where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in this answer is not.

The standard kernel theorem ([1] chapter 1, §1.3 pp. 11-20 and §3.5 pp. 73-79) guarantees "only" that there is a distribution which does the job, but does not guarantee that there exists an integral representation, though this can be true for general $n$-linear functionals on a particular function space, or for particular functionals on general function spaces: also, the integral representation may not have the form \eqref{1}. For example, in the case of linear functionals, an integral representation as a contour integral $$ f(\varphi)=\oint_\gamma v(z)\varphi(z)\mathrm{d}z\quad \varphi\in \mathscr{H\!\!o\!l}(D) $$ where

• $v(z)$ is the Fantappiè indicatrix of the functional, i.e. the (in general meromorphic) function $$ v(z)=f\left(\frac{1}{\zeta-z}\right)\quad(\zeta\text{ is the "inner" variable}) $$
• $\gamma\subset D\subseteq \Bbb C$ is a curve enclosing all singularities of $v$.

holds for every linear analytic functional $f\in\mathscr{O}(\Bbb C)$. And there are also more general "integral" representations (if you allow for the kernel $K$ to be a Radon Measure) which include large classes of linear functionals and more general generalized functions: see for example the paper [2].

References

[1] Gel’fand, I. M.; Vilenkin, N. Ya., Generalized functions. Vol. 4: Applications of harmonic analysis, Translated from the Russian by Amiel Feinstein. (English) New York and London: Academic Press. XIV, 384 p. (1964), MR0173945, Zbl 0136.11201.

[2] Kaneko, Akira, "Representation of hyperfunctions by measures and some of its applications", (English) Journal of the Faculty of Science, Section I A 19, 321-352 (1972), MR0336328, Zbl 0247.35007.

No, and a simple example goes as follows: $$ \begin{split} K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) & =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\ & \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N \end{split} $$ where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in this answer is not.

The standard kernel theorem ([1] chapter 1, §1.3 pp. 11-20 and §3.5 pp. 73-79) guarantees "only" that there is a distribution which does the job, but does not guarantee that there exists an integral representation, though this can be true for general $n$-linear functionals on particular function spaces.

References

Gel’fand, I. M.; Vilenkin, N. Ya., Generalized functions. Vol. 4: Applications of harmonic analysis, Translated from the Russian by Amiel Feinstein. (English) New York and London: Academic Press. XIV, 384 p. (1964), MR0173945, Zbl 0136.11201.

No, and a simple example goes as follows: $$ \begin{split} K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) & =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\ & \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N \end{split} $$ where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in this answer is not.

The standard kernel theorem ([1] chapter 1, §1.3 pp. 11-20 and §3.5 pp. 73-79) guarantees "only" that there is a distribution which does the job, but does not guarantee that there exists an integral representation, though this can be true for general $n$-linear functionals on a particular function space, or for particular functionals on general function spaces: also, the integral representation may not have the form \eqref{1}.

References

[1] Gel’fand, I. M.; Vilenkin, N. Ya., Generalized functions. Vol. 4: Applications of harmonic analysis, Translated from the Russian by Amiel Feinstein. (English) New York and London: Academic Press. XIV, 384 p. (1964), MR0173945, Zbl 0136.11201.