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The inf-convolution of the functions $f$ and $g$ defined on $\mathbb{R}^n$ is $$ h(x)=\inf _{y \in \mathbb{R}^n} f(y)+g(x-y) . $$ We can prove that if $f,g$ are convex functions, then $h$ is convex.

If $f(t) = \lVert t \rVert_1$, and $g(t) = \frac{1}{2}\lVert t \rVert_2^2$, how can we compute $h$?

My attempt: I try to calculate the derivative but $$ f(y) = \lVert y \rVert_1 =\sum_{i=1}^{n} |y_i|. $$ That isn't differentiable at $0$.

Thank you for your help

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    $\begingroup$ This question is not for this forum. You may want to try math.stackexchange.com $\endgroup$
    – Iosif Pinelis
    Commented Nov 13, 2023 at 2:07
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    $\begingroup$ What do you mean by $\Vert t\Vert_1$ and $\Vert t\Vert_2$? $\endgroup$
    – Piotr Hajlasz
    Commented Nov 13, 2023 at 4:55
  • $\begingroup$ Solve first in 1d where it is elementary and then the nd follows since $f(y)=\sum_i f_i(y_i)$. $\endgroup$
    – Giorgio Metafune
    Commented Nov 13, 2023 at 19:44
  • $\begingroup$ I mean that if $y=(y_1;y_2;....y_n)$, then $\lVert y \rVert_1 = \sum_{i=1}^{n} |y_i|$ and $\lVert y\rVert_2 = \sum_{i=1}^{n} \sqrt{y_1^2+y_2^2+...y_n^2}$ $\endgroup$
    – Pipnap
    Commented Nov 14, 2023 at 0:12

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