Simplex in a complex vector space

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I feel this may be a stupid question, but when I look up things on convexity, all definitions are in $mathbbR^n$. For example the definition of a simplex or Caratheodory's theorem. I can only find references of these in real spaces.



My question is: doesn't this also hold in complex spaces? I can't see why it wouldn't, but seeing as all definitions I find only mention the reals, I feel like I'm missing something.







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    up vote
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    favorite
    1












    I feel this may be a stupid question, but when I look up things on convexity, all definitions are in $mathbbR^n$. For example the definition of a simplex or Caratheodory's theorem. I can only find references of these in real spaces.



    My question is: doesn't this also hold in complex spaces? I can't see why it wouldn't, but seeing as all definitions I find only mention the reals, I feel like I'm missing something.







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      I feel this may be a stupid question, but when I look up things on convexity, all definitions are in $mathbbR^n$. For example the definition of a simplex or Caratheodory's theorem. I can only find references of these in real spaces.



      My question is: doesn't this also hold in complex spaces? I can't see why it wouldn't, but seeing as all definitions I find only mention the reals, I feel like I'm missing something.







      share|cite|improve this question












      I feel this may be a stupid question, but when I look up things on convexity, all definitions are in $mathbbR^n$. For example the definition of a simplex or Caratheodory's theorem. I can only find references of these in real spaces.



      My question is: doesn't this also hold in complex spaces? I can't see why it wouldn't, but seeing as all definitions I find only mention the reals, I feel like I'm missing something.









      share|cite|improve this question











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      share|cite|improve this question










      asked Aug 8 at 17:13









      user353840

      1217




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          We can extend the notion of convex sets to the complex domain without real worry if we keep our convex combinations $[x,y] = theta x + (1-theta)y : textreal 0leqthetaleq1$ to use real weightings. This corresponds to thinking about $mathbbC^n$ as its isomorphically isometric copy $mathbbR^2n$.



          The issue with extending things further to full generality lies with the lack of total ordering on the $mathbbC$. For example, the standard definition of convexity $f(theta x + (1-theta)y) leq theta f(x) + (1-theta) f(y)$ doesn't make much sense in a space like $mathbbC$ without a notion of '$leq$' that can compare every point.



          Nevertheless, people have defined a notion of 'cone' convex functions on an arbitrary vector space $f : mathcalVto mathcalW$ (such as $mathbbC^n$ or even the set of positive operators $mathbbS^n_+$ on $mathbbF^n$) as follows: If we have a convex cone $K$ on $mathcalW$, we define a partial ordering on $mathcalW$ so that its elements $apreceq_K b$ if and only if $b-ain K$. From this we can define $f$ to be $K$-convex if and only if
          $$
          f(theta x + (1-theta)y)preceq_K theta f(x) + (1-theta)f(y).
          $$
          It turns out that if $mathcalW$ is a Banach space many properties of regular convex functions (like the linear lower bound, which now uses Frechet derivatives, or the epigraph definition) carry over uninhibited. There are wonderful results in this area from matrix analysis. See Bhatia's text for more.






          share|cite|improve this answer




















          • Thanks! Seems like an interesting book, will definitely give it a closer look! So as long as I stick to convex sets, which applies to things like simplices and Caratheodory's theorem, I'm fine with working in $mathbbC$?
            – user353840
            Aug 8 at 18:07











          • @user353840 As far as I'm aware, yes. Just be cautious if you see a '$leq$' appear :)
            – cdipaolo
            Aug 8 at 20:49











          • I will, thank you!
            – user353840
            Aug 11 at 13:38










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          1 Answer
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          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          We can extend the notion of convex sets to the complex domain without real worry if we keep our convex combinations $[x,y] = theta x + (1-theta)y : textreal 0leqthetaleq1$ to use real weightings. This corresponds to thinking about $mathbbC^n$ as its isomorphically isometric copy $mathbbR^2n$.



          The issue with extending things further to full generality lies with the lack of total ordering on the $mathbbC$. For example, the standard definition of convexity $f(theta x + (1-theta)y) leq theta f(x) + (1-theta) f(y)$ doesn't make much sense in a space like $mathbbC$ without a notion of '$leq$' that can compare every point.



          Nevertheless, people have defined a notion of 'cone' convex functions on an arbitrary vector space $f : mathcalVto mathcalW$ (such as $mathbbC^n$ or even the set of positive operators $mathbbS^n_+$ on $mathbbF^n$) as follows: If we have a convex cone $K$ on $mathcalW$, we define a partial ordering on $mathcalW$ so that its elements $apreceq_K b$ if and only if $b-ain K$. From this we can define $f$ to be $K$-convex if and only if
          $$
          f(theta x + (1-theta)y)preceq_K theta f(x) + (1-theta)f(y).
          $$
          It turns out that if $mathcalW$ is a Banach space many properties of regular convex functions (like the linear lower bound, which now uses Frechet derivatives, or the epigraph definition) carry over uninhibited. There are wonderful results in this area from matrix analysis. See Bhatia's text for more.






          share|cite|improve this answer




















          • Thanks! Seems like an interesting book, will definitely give it a closer look! So as long as I stick to convex sets, which applies to things like simplices and Caratheodory's theorem, I'm fine with working in $mathbbC$?
            – user353840
            Aug 8 at 18:07











          • @user353840 As far as I'm aware, yes. Just be cautious if you see a '$leq$' appear :)
            – cdipaolo
            Aug 8 at 20:49











          • I will, thank you!
            – user353840
            Aug 11 at 13:38














          up vote
          2
          down vote



          accepted










          We can extend the notion of convex sets to the complex domain without real worry if we keep our convex combinations $[x,y] = theta x + (1-theta)y : textreal 0leqthetaleq1$ to use real weightings. This corresponds to thinking about $mathbbC^n$ as its isomorphically isometric copy $mathbbR^2n$.



          The issue with extending things further to full generality lies with the lack of total ordering on the $mathbbC$. For example, the standard definition of convexity $f(theta x + (1-theta)y) leq theta f(x) + (1-theta) f(y)$ doesn't make much sense in a space like $mathbbC$ without a notion of '$leq$' that can compare every point.



          Nevertheless, people have defined a notion of 'cone' convex functions on an arbitrary vector space $f : mathcalVto mathcalW$ (such as $mathbbC^n$ or even the set of positive operators $mathbbS^n_+$ on $mathbbF^n$) as follows: If we have a convex cone $K$ on $mathcalW$, we define a partial ordering on $mathcalW$ so that its elements $apreceq_K b$ if and only if $b-ain K$. From this we can define $f$ to be $K$-convex if and only if
          $$
          f(theta x + (1-theta)y)preceq_K theta f(x) + (1-theta)f(y).
          $$
          It turns out that if $mathcalW$ is a Banach space many properties of regular convex functions (like the linear lower bound, which now uses Frechet derivatives, or the epigraph definition) carry over uninhibited. There are wonderful results in this area from matrix analysis. See Bhatia's text for more.






          share|cite|improve this answer




















          • Thanks! Seems like an interesting book, will definitely give it a closer look! So as long as I stick to convex sets, which applies to things like simplices and Caratheodory's theorem, I'm fine with working in $mathbbC$?
            – user353840
            Aug 8 at 18:07











          • @user353840 As far as I'm aware, yes. Just be cautious if you see a '$leq$' appear :)
            – cdipaolo
            Aug 8 at 20:49











          • I will, thank you!
            – user353840
            Aug 11 at 13:38












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          We can extend the notion of convex sets to the complex domain without real worry if we keep our convex combinations $[x,y] = theta x + (1-theta)y : textreal 0leqthetaleq1$ to use real weightings. This corresponds to thinking about $mathbbC^n$ as its isomorphically isometric copy $mathbbR^2n$.



          The issue with extending things further to full generality lies with the lack of total ordering on the $mathbbC$. For example, the standard definition of convexity $f(theta x + (1-theta)y) leq theta f(x) + (1-theta) f(y)$ doesn't make much sense in a space like $mathbbC$ without a notion of '$leq$' that can compare every point.



          Nevertheless, people have defined a notion of 'cone' convex functions on an arbitrary vector space $f : mathcalVto mathcalW$ (such as $mathbbC^n$ or even the set of positive operators $mathbbS^n_+$ on $mathbbF^n$) as follows: If we have a convex cone $K$ on $mathcalW$, we define a partial ordering on $mathcalW$ so that its elements $apreceq_K b$ if and only if $b-ain K$. From this we can define $f$ to be $K$-convex if and only if
          $$
          f(theta x + (1-theta)y)preceq_K theta f(x) + (1-theta)f(y).
          $$
          It turns out that if $mathcalW$ is a Banach space many properties of regular convex functions (like the linear lower bound, which now uses Frechet derivatives, or the epigraph definition) carry over uninhibited. There are wonderful results in this area from matrix analysis. See Bhatia's text for more.






          share|cite|improve this answer












          We can extend the notion of convex sets to the complex domain without real worry if we keep our convex combinations $[x,y] = theta x + (1-theta)y : textreal 0leqthetaleq1$ to use real weightings. This corresponds to thinking about $mathbbC^n$ as its isomorphically isometric copy $mathbbR^2n$.



          The issue with extending things further to full generality lies with the lack of total ordering on the $mathbbC$. For example, the standard definition of convexity $f(theta x + (1-theta)y) leq theta f(x) + (1-theta) f(y)$ doesn't make much sense in a space like $mathbbC$ without a notion of '$leq$' that can compare every point.



          Nevertheless, people have defined a notion of 'cone' convex functions on an arbitrary vector space $f : mathcalVto mathcalW$ (such as $mathbbC^n$ or even the set of positive operators $mathbbS^n_+$ on $mathbbF^n$) as follows: If we have a convex cone $K$ on $mathcalW$, we define a partial ordering on $mathcalW$ so that its elements $apreceq_K b$ if and only if $b-ain K$. From this we can define $f$ to be $K$-convex if and only if
          $$
          f(theta x + (1-theta)y)preceq_K theta f(x) + (1-theta)f(y).
          $$
          It turns out that if $mathcalW$ is a Banach space many properties of regular convex functions (like the linear lower bound, which now uses Frechet derivatives, or the epigraph definition) carry over uninhibited. There are wonderful results in this area from matrix analysis. See Bhatia's text for more.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 8 at 17:35









          cdipaolo

          542211




          542211











          • Thanks! Seems like an interesting book, will definitely give it a closer look! So as long as I stick to convex sets, which applies to things like simplices and Caratheodory's theorem, I'm fine with working in $mathbbC$?
            – user353840
            Aug 8 at 18:07











          • @user353840 As far as I'm aware, yes. Just be cautious if you see a '$leq$' appear :)
            – cdipaolo
            Aug 8 at 20:49











          • I will, thank you!
            – user353840
            Aug 11 at 13:38
















          • Thanks! Seems like an interesting book, will definitely give it a closer look! So as long as I stick to convex sets, which applies to things like simplices and Caratheodory's theorem, I'm fine with working in $mathbbC$?
            – user353840
            Aug 8 at 18:07











          • @user353840 As far as I'm aware, yes. Just be cautious if you see a '$leq$' appear :)
            – cdipaolo
            Aug 8 at 20:49











          • I will, thank you!
            – user353840
            Aug 11 at 13:38















          Thanks! Seems like an interesting book, will definitely give it a closer look! So as long as I stick to convex sets, which applies to things like simplices and Caratheodory's theorem, I'm fine with working in $mathbbC$?
          – user353840
          Aug 8 at 18:07





          Thanks! Seems like an interesting book, will definitely give it a closer look! So as long as I stick to convex sets, which applies to things like simplices and Caratheodory's theorem, I'm fine with working in $mathbbC$?
          – user353840
          Aug 8 at 18:07













          @user353840 As far as I'm aware, yes. Just be cautious if you see a '$leq$' appear :)
          – cdipaolo
          Aug 8 at 20:49





          @user353840 As far as I'm aware, yes. Just be cautious if you see a '$leq$' appear :)
          – cdipaolo
          Aug 8 at 20:49













          I will, thank you!
          – user353840
          Aug 11 at 13:38




          I will, thank you!
          – user353840
          Aug 11 at 13:38












           

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