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10 votes
1 answer
900 views

Approximation of a compactly supported function by Gaussians

Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth function whose support is a closed interval, e.g. $\text{supp}(f)=[a,b]$. Then $f$ can be approximated (e.g. in $L^2$) by a linear combination of Gaussian ...
JohnA's user avatar
  • 710
5 votes
0 answers
471 views

A stronger Cauchy-Schwarz in infinite dimensional Hilbert spaces?

In this MSE and question and this MO question, stronger variants of the classical Cauchy-Schwarz inequality have been suggested in finite dimensional spaces. Can we find similar results for infinite ...
UserA's user avatar
  • 597
3 votes
3 answers
2k views

Determining if a set is a Basis for l^2

For each $ n\ge 1$ Define the vectors $e_n = (e_{nk})$ where $ k\ge 1$ and $ e_{nk} = \frac{1}{k^n}$ Is this set a basis for $l^2$? Thanks,
Ali's user avatar
  • 4,145
2 votes
0 answers
88 views

Dependence and $L^2$ projections of functions

tl;dr: Is it possible that the best approximation to a nonnegative function of three variables with a bivariate function is no better than the best univariate function? Let $w$ be a density on $\...
shawn532's user avatar
1 vote
1 answer
153 views

Optimal estimate in trace norm

Let $x,y$ be vectors of some Hilbert space of unit length. Then we can consider the projection $P_x:=\langle \bullet, x \rangle x$ and similarly $P_y.$ Assume then that we know that $\left\lVert x-...
Xing Wang's user avatar
  • 119
1 vote
1 answer
112 views

Orthogonal complement vector space

Let $X$ be a vector space contained in $H^{1}(\mathbb R^d),$ then we can study $X^{\perp_{L^2}}:=\left\{ \xi \in L^2; \langle \xi, x \rangle_{L^2} =0 \ \forall x \in X \right\}$ and $X^{\perp_{H^{-...
Ulan12's user avatar
  • 13
0 votes
1 answer
269 views

Determine if an integral expression is in $L^2(\mathbb{R})$

Note: This is a simplified version of the following question. I did not get a full response and realized can make it simpler to have my main interrogation answered. I decided to write it as a ...
Gateau au fromage's user avatar