Solution verification post : Problem $12.1$, Mathematical Analysis, Apostol

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As a continuation of this post, I'm starting to post my solutions (or attempts) to the exercise problems of Chapter 12 (Multivariable Differential Calculus), Mathematical Analysis by Apostol (suggested by Masacroso). Since I'm essentially self-studying, I'd really appreciate if anyone checks the solutions and let me know if there is any gap in my arguments or if there exists any cleverer solutions. Thank you.




Problem $12.1.$ Let $S$ be an open subset of $mathbbR^n$, and let $f:S to mathbbR$ be a real-valued function with finite partial derivative $D_1f,dots,D_nf$ on $S$. If $f$ has a local maximum or a local minimum at a point $c$ in $S$, prove that $D_kfleft(cright)=0 ,,forall ,k$.




Solution. Suppose that $f$ has a local maximum at $c in S$. This implies that $exists delta>0$ such that
$$f(c) geq f(x) text for all x in B_delta(c)=lefty in S : leftVert y-c rightVert < delta right tag1$$
Let $k in 1,dots,n$. Now,
$$D_kf(c)=lim_h to 0 fracf(c+he_k)-f(c)h tag2$$
It is given that $(2)$ exists finitely, then
$$D_kf(c)=lim_h to 0;,h in (0,delta) fracf(c+he_k)-f(c)h leq 0 tag3$$
On the other hand,
$$D_kf(c)=lim_h to 0;,h in (-delta,0) fracf(c+he_k)-f(c)h geq 0 tag4$$
$(3)$ and $(4)$ $Rightarrow$ $D_kf(c)=0$. This holds for every $k in 1,dots,n$ and hence we are done.










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    up vote
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    down vote

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    As a continuation of this post, I'm starting to post my solutions (or attempts) to the exercise problems of Chapter 12 (Multivariable Differential Calculus), Mathematical Analysis by Apostol (suggested by Masacroso). Since I'm essentially self-studying, I'd really appreciate if anyone checks the solutions and let me know if there is any gap in my arguments or if there exists any cleverer solutions. Thank you.




    Problem $12.1.$ Let $S$ be an open subset of $mathbbR^n$, and let $f:S to mathbbR$ be a real-valued function with finite partial derivative $D_1f,dots,D_nf$ on $S$. If $f$ has a local maximum or a local minimum at a point $c$ in $S$, prove that $D_kfleft(cright)=0 ,,forall ,k$.




    Solution. Suppose that $f$ has a local maximum at $c in S$. This implies that $exists delta>0$ such that
    $$f(c) geq f(x) text for all x in B_delta(c)=lefty in S : leftVert y-c rightVert < delta right tag1$$
    Let $k in 1,dots,n$. Now,
    $$D_kf(c)=lim_h to 0 fracf(c+he_k)-f(c)h tag2$$
    It is given that $(2)$ exists finitely, then
    $$D_kf(c)=lim_h to 0;,h in (0,delta) fracf(c+he_k)-f(c)h leq 0 tag3$$
    On the other hand,
    $$D_kf(c)=lim_h to 0;,h in (-delta,0) fracf(c+he_k)-f(c)h geq 0 tag4$$
    $(3)$ and $(4)$ $Rightarrow$ $D_kf(c)=0$. This holds for every $k in 1,dots,n$ and hence we are done.










    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      As a continuation of this post, I'm starting to post my solutions (or attempts) to the exercise problems of Chapter 12 (Multivariable Differential Calculus), Mathematical Analysis by Apostol (suggested by Masacroso). Since I'm essentially self-studying, I'd really appreciate if anyone checks the solutions and let me know if there is any gap in my arguments or if there exists any cleverer solutions. Thank you.




      Problem $12.1.$ Let $S$ be an open subset of $mathbbR^n$, and let $f:S to mathbbR$ be a real-valued function with finite partial derivative $D_1f,dots,D_nf$ on $S$. If $f$ has a local maximum or a local minimum at a point $c$ in $S$, prove that $D_kfleft(cright)=0 ,,forall ,k$.




      Solution. Suppose that $f$ has a local maximum at $c in S$. This implies that $exists delta>0$ such that
      $$f(c) geq f(x) text for all x in B_delta(c)=lefty in S : leftVert y-c rightVert < delta right tag1$$
      Let $k in 1,dots,n$. Now,
      $$D_kf(c)=lim_h to 0 fracf(c+he_k)-f(c)h tag2$$
      It is given that $(2)$ exists finitely, then
      $$D_kf(c)=lim_h to 0;,h in (0,delta) fracf(c+he_k)-f(c)h leq 0 tag3$$
      On the other hand,
      $$D_kf(c)=lim_h to 0;,h in (-delta,0) fracf(c+he_k)-f(c)h geq 0 tag4$$
      $(3)$ and $(4)$ $Rightarrow$ $D_kf(c)=0$. This holds for every $k in 1,dots,n$ and hence we are done.










      share|cite|improve this question













      As a continuation of this post, I'm starting to post my solutions (or attempts) to the exercise problems of Chapter 12 (Multivariable Differential Calculus), Mathematical Analysis by Apostol (suggested by Masacroso). Since I'm essentially self-studying, I'd really appreciate if anyone checks the solutions and let me know if there is any gap in my arguments or if there exists any cleverer solutions. Thank you.




      Problem $12.1.$ Let $S$ be an open subset of $mathbbR^n$, and let $f:S to mathbbR$ be a real-valued function with finite partial derivative $D_1f,dots,D_nf$ on $S$. If $f$ has a local maximum or a local minimum at a point $c$ in $S$, prove that $D_kfleft(cright)=0 ,,forall ,k$.




      Solution. Suppose that $f$ has a local maximum at $c in S$. This implies that $exists delta>0$ such that
      $$f(c) geq f(x) text for all x in B_delta(c)=lefty in S : leftVert y-c rightVert < delta right tag1$$
      Let $k in 1,dots,n$. Now,
      $$D_kf(c)=lim_h to 0 fracf(c+he_k)-f(c)h tag2$$
      It is given that $(2)$ exists finitely, then
      $$D_kf(c)=lim_h to 0;,h in (0,delta) fracf(c+he_k)-f(c)h leq 0 tag3$$
      On the other hand,
      $$D_kf(c)=lim_h to 0;,h in (-delta,0) fracf(c+he_k)-f(c)h geq 0 tag4$$
      $(3)$ and $(4)$ $Rightarrow$ $D_kf(c)=0$. This holds for every $k in 1,dots,n$ and hence we are done.







      real-analysis analysis multivariable-calculus derivatives proof-verification






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      asked Sep 10 at 13:03









      Eugenia

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          Yes, your solution is correct. The same argument is used in the one-dimensional case. In fact, you could reduce your problem to the one-dimensional case by considering
          $$g_k(t):= f(a+t e_k)$$
          in a neighboorhoud of $0$ (exists because $S$ is open) and fixed $a in S$. Here $e_1,ldots,e_n$ denote the 'standard basis' of $mathbbR^n$. Assuming that $f$ has a local maximum in $a in S$ implies that $g_1,ldots,g_n$ have a local maximum in $t=0$. Thus we have
          $$0=g'_k(0) = left. fracmathrmdmathrmd x_k f(x) right|_x=a.$$






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            Yes, your solution is correct. The same argument is used in the one-dimensional case. In fact, you could reduce your problem to the one-dimensional case by considering
            $$g_k(t):= f(a+t e_k)$$
            in a neighboorhoud of $0$ (exists because $S$ is open) and fixed $a in S$. Here $e_1,ldots,e_n$ denote the 'standard basis' of $mathbbR^n$. Assuming that $f$ has a local maximum in $a in S$ implies that $g_1,ldots,g_n$ have a local maximum in $t=0$. Thus we have
            $$0=g'_k(0) = left. fracmathrmdmathrmd x_k f(x) right|_x=a.$$






            share|cite|improve this answer
























              up vote
              0
              down vote













              Yes, your solution is correct. The same argument is used in the one-dimensional case. In fact, you could reduce your problem to the one-dimensional case by considering
              $$g_k(t):= f(a+t e_k)$$
              in a neighboorhoud of $0$ (exists because $S$ is open) and fixed $a in S$. Here $e_1,ldots,e_n$ denote the 'standard basis' of $mathbbR^n$. Assuming that $f$ has a local maximum in $a in S$ implies that $g_1,ldots,g_n$ have a local maximum in $t=0$. Thus we have
              $$0=g'_k(0) = left. fracmathrmdmathrmd x_k f(x) right|_x=a.$$






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                Yes, your solution is correct. The same argument is used in the one-dimensional case. In fact, you could reduce your problem to the one-dimensional case by considering
                $$g_k(t):= f(a+t e_k)$$
                in a neighboorhoud of $0$ (exists because $S$ is open) and fixed $a in S$. Here $e_1,ldots,e_n$ denote the 'standard basis' of $mathbbR^n$. Assuming that $f$ has a local maximum in $a in S$ implies that $g_1,ldots,g_n$ have a local maximum in $t=0$. Thus we have
                $$0=g'_k(0) = left. fracmathrmdmathrmd x_k f(x) right|_x=a.$$






                share|cite|improve this answer












                Yes, your solution is correct. The same argument is used in the one-dimensional case. In fact, you could reduce your problem to the one-dimensional case by considering
                $$g_k(t):= f(a+t e_k)$$
                in a neighboorhoud of $0$ (exists because $S$ is open) and fixed $a in S$. Here $e_1,ldots,e_n$ denote the 'standard basis' of $mathbbR^n$. Assuming that $f$ has a local maximum in $a in S$ implies that $g_1,ldots,g_n$ have a local maximum in $t=0$. Thus we have
                $$0=g'_k(0) = left. fracmathrmdmathrmd x_k f(x) right|_x=a.$$







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Sep 10 at 13:13









                p4sch

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