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6 votes
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Nature of the Euclidean Norm

This is subtle. In mathematics we have the freedom to define distances however we want depending on the application, and for example sometimes rather than the Euclidean norm we use what are called the ...
Qiaochu Yuan's user avatar
3 votes
Accepted

For a given $k \in \mathbb{N}$, define the mapping $A: X \to X$ by the rule $ (Af)(x) = x^k f(x). $

To show that $A$ is linear means to show that $$ A(\alpha f+\beta g)=\alpha Af+\beta Ag $$ or, equivalently, that $$ x^k(\alpha f(x)+\beta g(x))=\alpha x^kf(x)+\beta x^kg(x). $$ This identity clearly ...
John B's user avatar
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3 votes

Metric on $\mathbb R^{m \times n}$ based on the Frobenius norm

$\def\R{\mathbb{R}}\def\T{^{\mathrm{T}}}\def\ge{\geqslant}\def\le{\leqslant}\def\peq{\mathrel{\phantom{=}}}\def\norm#1{\left\lVert#1\right\rVert}\def\paren#1{\left(#1\right)}$Proposition 1 (Ptolemy): ...
Ѕᴀᴀᴅ's user avatar
  • 34.6k
2 votes

Metric on $\mathbb R^{m \times n}$ based on the Frobenius norm

The following is a partial answer as it only fully addresses your second question. As for the first question: while numerics suggest that your map $d$ is indeed a metric, I cannot even prove the ...
Frederik vom Ende's user avatar
1 vote

Conceptual difference between regularized and constrained optimization.

I think that conceptually the key difference is very direct - the first formulation does not impose an explicit constraint on the maximum allowable fitting error, but the second does. Regarding the ...
Alex Shtoff's user avatar
1 vote

Nature of the Euclidean Norm

The definition is partly "an greed-upon convention", but at the same tame there is a reason to define it the way it is. There is mathematical notion called "metric" or "...
César VB's user avatar
  • 495
1 vote
Accepted

I want to prove that $T$ belongs to $B(X)$, meaning it is a bounded linear operator.

I think that $f \in X^*$ should be a bounded linear operator as well. I'm assuming it is a operator of this kind: $f:X\to\mathbb{K}$ where $\mathbb{K}$ is a field equal to $\mathbb{R}$ or $\mathbb{C}$...
Alejandro Sánchez Yalí's user avatar

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