is $f(x,y)=e^{x-y}$ globallyconvex?












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The function is positive $forall (x,y)in R^2$.If I consider the restriction to y-axis $f(0,y)=e^{-y}rightarrow 0^+$ for $yrightarrow +infty$ so Inf f=$0$.
To study convexity can I consider the matrix Hessian?






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    1












    $begingroup$


    The function is positive $forall (x,y)in R^2$.If I consider the restriction to y-axis $f(0,y)=e^{-y}rightarrow 0^+$ for $yrightarrow +infty$ so Inf f=$0$.
    To study convexity can I consider the matrix Hessian?






    real-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      The function is positive $forall (x,y)in R^2$.If I consider the restriction to y-axis $f(0,y)=e^{-y}rightarrow 0^+$ for $yrightarrow +infty$ so Inf f=$0$.
      To study convexity can I consider the matrix Hessian?






      real-analysis







      share|cite|improve this question









      $endgroup$




      The function is positive $forall (x,y)in R^2$.If I consider the restriction to y-axis $f(0,y)=e^{-y}rightarrow 0^+$ for $yrightarrow +infty$ so Inf f=$0$.
      To study convexity can I consider the matrix Hessian?






      real-analysis




      real-analysis






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      share|cite|improve this question










      asked Dec 4 '18 at 15:14









      Giulia B.Giulia B.

      415211




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          3 Answers
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          $begingroup$

          For convexity I would start by changing coordiantes from $(x,y)$ to $(p,q)=(x-y,x+y)$. Then $f(p,q)=e^p$, and it is easy to directly compare $f(v)$ and $f(w)$ with $f(tv+(1-t)w)$. One then sees that the function inherits convexity from the real exponential function.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            If $p = (p_1,p_2)$ and $q = (q_1,q_2)$ and $0 le t le 1$,
            $$f(t p + (1-t) q) = exp(t(p_1 - p_2)+(1-t)(q_1-q_2)) le
            t exp(p_1-p_2) + (1-t) exp(q_1 - q_2) = t f(p) + (1-t) f(q)$$
            by convexity of $exp$.



            Somewhat more generally, if $g$ is an affine map and $h$ is convex then $h circ g$ is convex.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The function is globally convex but hasn't absolute minimum?
              $endgroup$
              – Giulia B.
              Dec 4 '18 at 15:24










            • $begingroup$
              Yes. Why is that surprising to you?
              $endgroup$
              – Robert Israel
              Dec 4 '18 at 15:25










            • $begingroup$
              Yes: for precisely the same reason that $x mapsto e^x$ is globally convex but has no absolute minimum.
              $endgroup$
              – user3482749
              Dec 4 '18 at 15:26



















            1












            $begingroup$

            Take any $(x,y), (z,w)inmathbb{R}^2$, and any $t in mathbb{R}$. Then $f(t(x,y)+(1-t)(z,w)) = e^{tx+(1-t)z - ty - (1-t)w} = e^{t(x-y)+(1-t)(z-w)}$, and $tf(x,y)+(1-t)f(z,w) = te^{x-y}+(1-t)e^{z-w}$. We need to show that the former is never greater than the latter.



            But $g: mathbb{R}tomathbb{R}: x mapsto e^x$ is convex, so we have (taking our points in the definition of the convexity of $g$ to be $x-y$ and $z-w$)



            $$e^{t(x-y)+(1-t)(z-w)} leq te^{x-y}+(1-t)e^{z-w},$$



            But the left-hand-side of this is exactly $f(t(x,y)+(1-t)(z,w))$, and the right-hand-side is exactly $tf(x,y)+(1-t)f(z,w)$, so indeed, $f$ is convex.






            share|cite|improve this answer









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              3 Answers
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              3 Answers
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              1












              $begingroup$

              For convexity I would start by changing coordiantes from $(x,y)$ to $(p,q)=(x-y,x+y)$. Then $f(p,q)=e^p$, and it is easy to directly compare $f(v)$ and $f(w)$ with $f(tv+(1-t)w)$. One then sees that the function inherits convexity from the real exponential function.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                For convexity I would start by changing coordiantes from $(x,y)$ to $(p,q)=(x-y,x+y)$. Then $f(p,q)=e^p$, and it is easy to directly compare $f(v)$ and $f(w)$ with $f(tv+(1-t)w)$. One then sees that the function inherits convexity from the real exponential function.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For convexity I would start by changing coordiantes from $(x,y)$ to $(p,q)=(x-y,x+y)$. Then $f(p,q)=e^p$, and it is easy to directly compare $f(v)$ and $f(w)$ with $f(tv+(1-t)w)$. One then sees that the function inherits convexity from the real exponential function.






                  share|cite|improve this answer









                  $endgroup$



                  For convexity I would start by changing coordiantes from $(x,y)$ to $(p,q)=(x-y,x+y)$. Then $f(p,q)=e^p$, and it is easy to directly compare $f(v)$ and $f(w)$ with $f(tv+(1-t)w)$. One then sees that the function inherits convexity from the real exponential function.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 15:19









                  Henning MakholmHenning Makholm

                  239k17304541




                  239k17304541























                      1












                      $begingroup$

                      If $p = (p_1,p_2)$ and $q = (q_1,q_2)$ and $0 le t le 1$,
                      $$f(t p + (1-t) q) = exp(t(p_1 - p_2)+(1-t)(q_1-q_2)) le
                      t exp(p_1-p_2) + (1-t) exp(q_1 - q_2) = t f(p) + (1-t) f(q)$$
                      by convexity of $exp$.



                      Somewhat more generally, if $g$ is an affine map and $h$ is convex then $h circ g$ is convex.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        The function is globally convex but hasn't absolute minimum?
                        $endgroup$
                        – Giulia B.
                        Dec 4 '18 at 15:24










                      • $begingroup$
                        Yes. Why is that surprising to you?
                        $endgroup$
                        – Robert Israel
                        Dec 4 '18 at 15:25










                      • $begingroup$
                        Yes: for precisely the same reason that $x mapsto e^x$ is globally convex but has no absolute minimum.
                        $endgroup$
                        – user3482749
                        Dec 4 '18 at 15:26
















                      1












                      $begingroup$

                      If $p = (p_1,p_2)$ and $q = (q_1,q_2)$ and $0 le t le 1$,
                      $$f(t p + (1-t) q) = exp(t(p_1 - p_2)+(1-t)(q_1-q_2)) le
                      t exp(p_1-p_2) + (1-t) exp(q_1 - q_2) = t f(p) + (1-t) f(q)$$
                      by convexity of $exp$.



                      Somewhat more generally, if $g$ is an affine map and $h$ is convex then $h circ g$ is convex.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        The function is globally convex but hasn't absolute minimum?
                        $endgroup$
                        – Giulia B.
                        Dec 4 '18 at 15:24










                      • $begingroup$
                        Yes. Why is that surprising to you?
                        $endgroup$
                        – Robert Israel
                        Dec 4 '18 at 15:25










                      • $begingroup$
                        Yes: for precisely the same reason that $x mapsto e^x$ is globally convex but has no absolute minimum.
                        $endgroup$
                        – user3482749
                        Dec 4 '18 at 15:26














                      1












                      1








                      1





                      $begingroup$

                      If $p = (p_1,p_2)$ and $q = (q_1,q_2)$ and $0 le t le 1$,
                      $$f(t p + (1-t) q) = exp(t(p_1 - p_2)+(1-t)(q_1-q_2)) le
                      t exp(p_1-p_2) + (1-t) exp(q_1 - q_2) = t f(p) + (1-t) f(q)$$
                      by convexity of $exp$.



                      Somewhat more generally, if $g$ is an affine map and $h$ is convex then $h circ g$ is convex.






                      share|cite|improve this answer









                      $endgroup$



                      If $p = (p_1,p_2)$ and $q = (q_1,q_2)$ and $0 le t le 1$,
                      $$f(t p + (1-t) q) = exp(t(p_1 - p_2)+(1-t)(q_1-q_2)) le
                      t exp(p_1-p_2) + (1-t) exp(q_1 - q_2) = t f(p) + (1-t) f(q)$$
                      by convexity of $exp$.



                      Somewhat more generally, if $g$ is an affine map and $h$ is convex then $h circ g$ is convex.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 4 '18 at 15:22









                      Robert IsraelRobert Israel

                      320k23210462




                      320k23210462












                      • $begingroup$
                        The function is globally convex but hasn't absolute minimum?
                        $endgroup$
                        – Giulia B.
                        Dec 4 '18 at 15:24










                      • $begingroup$
                        Yes. Why is that surprising to you?
                        $endgroup$
                        – Robert Israel
                        Dec 4 '18 at 15:25










                      • $begingroup$
                        Yes: for precisely the same reason that $x mapsto e^x$ is globally convex but has no absolute minimum.
                        $endgroup$
                        – user3482749
                        Dec 4 '18 at 15:26


















                      • $begingroup$
                        The function is globally convex but hasn't absolute minimum?
                        $endgroup$
                        – Giulia B.
                        Dec 4 '18 at 15:24










                      • $begingroup$
                        Yes. Why is that surprising to you?
                        $endgroup$
                        – Robert Israel
                        Dec 4 '18 at 15:25










                      • $begingroup$
                        Yes: for precisely the same reason that $x mapsto e^x$ is globally convex but has no absolute minimum.
                        $endgroup$
                        – user3482749
                        Dec 4 '18 at 15:26
















                      $begingroup$
                      The function is globally convex but hasn't absolute minimum?
                      $endgroup$
                      – Giulia B.
                      Dec 4 '18 at 15:24




                      $begingroup$
                      The function is globally convex but hasn't absolute minimum?
                      $endgroup$
                      – Giulia B.
                      Dec 4 '18 at 15:24












                      $begingroup$
                      Yes. Why is that surprising to you?
                      $endgroup$
                      – Robert Israel
                      Dec 4 '18 at 15:25




                      $begingroup$
                      Yes. Why is that surprising to you?
                      $endgroup$
                      – Robert Israel
                      Dec 4 '18 at 15:25












                      $begingroup$
                      Yes: for precisely the same reason that $x mapsto e^x$ is globally convex but has no absolute minimum.
                      $endgroup$
                      – user3482749
                      Dec 4 '18 at 15:26




                      $begingroup$
                      Yes: for precisely the same reason that $x mapsto e^x$ is globally convex but has no absolute minimum.
                      $endgroup$
                      – user3482749
                      Dec 4 '18 at 15:26











                      1












                      $begingroup$

                      Take any $(x,y), (z,w)inmathbb{R}^2$, and any $t in mathbb{R}$. Then $f(t(x,y)+(1-t)(z,w)) = e^{tx+(1-t)z - ty - (1-t)w} = e^{t(x-y)+(1-t)(z-w)}$, and $tf(x,y)+(1-t)f(z,w) = te^{x-y}+(1-t)e^{z-w}$. We need to show that the former is never greater than the latter.



                      But $g: mathbb{R}tomathbb{R}: x mapsto e^x$ is convex, so we have (taking our points in the definition of the convexity of $g$ to be $x-y$ and $z-w$)



                      $$e^{t(x-y)+(1-t)(z-w)} leq te^{x-y}+(1-t)e^{z-w},$$



                      But the left-hand-side of this is exactly $f(t(x,y)+(1-t)(z,w))$, and the right-hand-side is exactly $tf(x,y)+(1-t)f(z,w)$, so indeed, $f$ is convex.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Take any $(x,y), (z,w)inmathbb{R}^2$, and any $t in mathbb{R}$. Then $f(t(x,y)+(1-t)(z,w)) = e^{tx+(1-t)z - ty - (1-t)w} = e^{t(x-y)+(1-t)(z-w)}$, and $tf(x,y)+(1-t)f(z,w) = te^{x-y}+(1-t)e^{z-w}$. We need to show that the former is never greater than the latter.



                        But $g: mathbb{R}tomathbb{R}: x mapsto e^x$ is convex, so we have (taking our points in the definition of the convexity of $g$ to be $x-y$ and $z-w$)



                        $$e^{t(x-y)+(1-t)(z-w)} leq te^{x-y}+(1-t)e^{z-w},$$



                        But the left-hand-side of this is exactly $f(t(x,y)+(1-t)(z,w))$, and the right-hand-side is exactly $tf(x,y)+(1-t)f(z,w)$, so indeed, $f$ is convex.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Take any $(x,y), (z,w)inmathbb{R}^2$, and any $t in mathbb{R}$. Then $f(t(x,y)+(1-t)(z,w)) = e^{tx+(1-t)z - ty - (1-t)w} = e^{t(x-y)+(1-t)(z-w)}$, and $tf(x,y)+(1-t)f(z,w) = te^{x-y}+(1-t)e^{z-w}$. We need to show that the former is never greater than the latter.



                          But $g: mathbb{R}tomathbb{R}: x mapsto e^x$ is convex, so we have (taking our points in the definition of the convexity of $g$ to be $x-y$ and $z-w$)



                          $$e^{t(x-y)+(1-t)(z-w)} leq te^{x-y}+(1-t)e^{z-w},$$



                          But the left-hand-side of this is exactly $f(t(x,y)+(1-t)(z,w))$, and the right-hand-side is exactly $tf(x,y)+(1-t)f(z,w)$, so indeed, $f$ is convex.






                          share|cite|improve this answer









                          $endgroup$



                          Take any $(x,y), (z,w)inmathbb{R}^2$, and any $t in mathbb{R}$. Then $f(t(x,y)+(1-t)(z,w)) = e^{tx+(1-t)z - ty - (1-t)w} = e^{t(x-y)+(1-t)(z-w)}$, and $tf(x,y)+(1-t)f(z,w) = te^{x-y}+(1-t)e^{z-w}$. We need to show that the former is never greater than the latter.



                          But $g: mathbb{R}tomathbb{R}: x mapsto e^x$ is convex, so we have (taking our points in the definition of the convexity of $g$ to be $x-y$ and $z-w$)



                          $$e^{t(x-y)+(1-t)(z-w)} leq te^{x-y}+(1-t)e^{z-w},$$



                          But the left-hand-side of this is exactly $f(t(x,y)+(1-t)(z,w))$, and the right-hand-side is exactly $tf(x,y)+(1-t)f(z,w)$, so indeed, $f$ is convex.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 4 '18 at 15:25









                          user3482749user3482749

                          4,057618




                          4,057618






























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