It is instructive to perform the p and q integrals in the above formula first, which has the effect of computing the ordinary Fourier transform ~ of the function , while leaving the operator (+). The transform can be written as the sum of elements of $\mathbf{s}$ and $\mathbf{x}$ as: :�LJ�Tq5��c�j���D�n�k��C= /Length 2324 $$f(\mathbf{x})=a\left\Vert \mathbf{x} - \b{y} \right\Vert _{1}$$ That in turn, means that the supremum is found for $x_{i}=y_i$, which leads to: \end{cases}$$ the Legendre transform is $$\sup_{\mathbf{x}} \mathbf{s}^{T}\mathbf{x}-\frac{1}{2}\left\Vert \b{A}(\mathbf{y}-\mathbf{x})\right\Vert_{2}^{2}$$ In it, the slope of the tangent ($s$) times the horizontal side ($x^\star$) equals the vertical side, which is the sum of $f(x)$ and $f^{\star}(s)$.
Consider the case $f(x)=|x|$. $$\b{s}^{T}\left(\b{A}^T \b{A}\right)^{-1}\b{s}+\b{s}^{T}\b{y}-\frac{1}{2}\mathbf{s}^{T}\left(\b{A}^T \b{A}\right)^{-1}\mathbf{s}$$ This way is easy to see that if any element $s_{i}>a$, then any vector $\mathbf{x}$ with $x_{i}\rightarrow\infty$ will push the transform to infinity. ���y�8Hz�,�
���Y�����N&�"1�z�z~�� l�'6��'��X@Xl�~��M��t@N1��n�f"S� ܮ'�W����e�q2b"N��(�#�*��o@?�����k�M���Z@,�U��ȡ4��/Oڬ���J_�fi��}�@k�Y��uNiL�����:�����u0E�8�� R1��A�/ �&itN�q���ۆ�� The Legendre transform is pretty useful on its own, but it is limited to convex and differentiable functions. The Legendre transform is an encoding of the convex hull of a function's epigraph in terms of it's supporting hyperplanes. We know that at $\mathbf{x}^{\star}$ the tangent touches $f(\mathbf{x})$, therefore: The weighted least squares problem is defined as: Of course there are many ways to get a function of $\mathbf{s}$, not necessarily the one presented right now. $$f(\mathbf{x})=\frac{1}{2}\left\Vert \b{A} (\b{y}-\b{x})\right\Vert _{2}^{2}$$ If we want a function of $\mathbf{s}$, we could first find the inverse of $\mathbf{s}(\mathbf{x})$, say $\mathbf{x}(\mathbf{s})$, and apply it to $f(\mathbf{x})$. $$\sup\sum\left(s_{i}x_{i}-a\left|x_{i} - y_i\right|\right)=\sup\sum\left(\begin{cases} +\infty & \mbox{otherwise} ��a�F��? "*x�nu��"7����݆� ��Xw�%wTe���6��h�Ӳ;0�wEV����D�ē*ut�5�D�b�5��*O6&��xg}$W�q'��rhJ5,����yF'h�֠M�������(,Z�S.�����:֢?=Wtz����Y�xwdt����@c���f�b3\F5��V�k�V, In mathematics and mathematical optimization, the convex conjugate of a function is a generalization of the Legendre transformation which applies to non-convex functions. stream xI�d!y6��Q�,���Co.b�C�%$z�}�eH���f�����3 [ �!�i�8ݜ�����r�M�?,�a�'\0$�}��6���k���^aC�¹X���_(K3dś�榡R]�9�L"r�I���M�K(��؈�����2��C������S��>�E�a�� !�IO���y�c��y��1Q��_��:�*�Q1��r�R����y^=U�UgrU�I��Ӊ����>A+�/_��xF�O�>X���/�� ��H����vΫl��}���ۀ�s|_���섗���x��x�CP,�L���$M��A��s�E�W�|������ Let's call the term $-b$, $f^{\star}(\mathbf{s})$. $$\mathbf{s}=\left.\frac{df(\mathbf{x})}{d\mathbf{x}}\right|_{\mathbf{x}=\mathbf{x}^{\star}}$$ \end{cases}\right)$$ ���]��U�L_<4��!���e��8�KWA���A�C �9^�PP�CZ\dޒg��9,ᙞ8X��TQ��r�k��9�Bg:��R/� !��.�#�/i�P���T����R�͊��'�o��}ɀ�����X��ɚC!khp��&u���|�PW��H4Q��,�C%��n��'���`�+�0�7���;}6��/)��
If I helped you in some way, please help me back by liking this website on the bottom of the page or clicking on the link below. f^{\star}(\mathbf{0}) & = &\mathbf{0}^{T}\mathbf{x}_{min}-f(\mathbf{x}_{min}) \\ & = & f(\mathbf{x}) where $b$ represents the value of the tangent when $\mathbf{x=0}$. For any $-a\le s_{i}\le a$, $\left(s_{i}-a\right)\le 0$ and $\left(s_{i}+a\right)\ge 0$. When $\b{y = 0}$, the above expression reverts to the $\ell_1$-norm case. $$y=\mathbf{s}^{T}\mathbf{x}+b,$$ Instead, the Legendre transform is the expression: Therefore: WF�S�B�݀. $$\frac{1}{2}\left\Vert \mathbf{s}\right\Vert _{2}^{2}+\mathbf{s}^T\mathbf{y}$$ The Legendre transform exploits a special feature of a convex (or concave) function f(x): its slope f0(x) is monotonic and hence is a single-valued and invertible function of x. Now we replace $\mathbf{x}$ in the original transform: In this sense, it resembles (geometric) du-ality transformations. The common case when $\b{A}$ is a diagonal matrix $\b{\Lambda}$, the above expression can be simplified: $$\sup_{x_i} (s_i x_i - a|x_i - y_i| ) = \begin{cases} >> Let's go into the math. h�1I�i���];�8� ܐ��������pD! $$f^{\star}(\mathbf{s}) = \mathbf{s}^{T}\mathbf{x}-f(\mathbf{x})$$ $$f^\star(\mathbf{s})=\mathbf{s}^{T}\mathbf{x}(\mathbf{s})-f(\mathbf{x}(\mathbf{s}))$$ In a geometric sense, there is a triangle shown. \b{s}^T\b{y} & \left|s_{i}\right|\le a,\forall i\\ $$\begin{eqnarray*} We would then get $f(\mathbf{x}(\mathbf{s}))$. $$f(\mathbf{x}^{\star}) = \mathbf{s}^{T}\mathbf{x}^{\star}+b$$ In words, for any point $\mathbf{x}=\mathbf{x}^{\star}$, the transform is the negative of parameter $b$ of the tangent line at that point (the value of the tangent line when $\mathbf{x}=\mathbf{0}$). To see how it works, you must first realize that for a convex and smooth function there is a one-to-one correspondence between $\mathbf{x}$ and $\mathbf{s}(\mathbf{x})$, because $\mathbf{s}(\mathbf{x})$ is a monotonic function of $\mathbf{x}$, i.e., the derivative is always increasing with $\mathbf{x}$. \left(s_{i}+a\right)x_{i} -a\cdot y_i & ,x_{i} - y_i \lt 0 Pretty cool, uh? If any of these properties fail, the transform cannot be used. $$\frac{1}{2}\b{s}^{T}\left(\b{A}^T \b{A}\right)^{-1}\b{s}+\b{s}^{T}\b{y}$$ Suffice to say that at the minimum, the slope of the tangent is zero. In the same way that the slope of $f(\mathbf{x})$ is $\mathbf{s}$, we have that the slope of $f^{\star}(\mathbf{s})$ is $\mathbf{x}$: The Weyl transform (or Weyl quantization) of the function f is given by the following operator in Hilbert space, [] = ∬ ∬ (,) (((−) + (−))).
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