Show that the two norms in $E=left{fin mathcal{C}^1[0,1]:f(0)=0right}$ are equivalent?












5












$begingroup$


I can't show that the two norms defined as
$$||f||_{infty}=sup_{xin[0,1]}|f(x)+ f'(x)|$$
and
$$N(f)=sup_{xin[0,1]}|f(x)| + sup_{xin[0,1]}|f'(x)|$$
are equivalent in $E={finmathcal{C}^1([0,1]) text{ s.t. } f(0)=0}$.



A first inequality in one sense is trivial.



For the other inequality, I can write:



$$
f(x)=f(0)+int_0^x f'(t) dt
$$
Which implies
$$
sup_{xin[0,1]}|f(x)| leq sup_{xin[0,1]}|f'(x)| .
$$
And then
$$
N(f)leq 2 sup_{xin[0,1]}|f'(x)| .
$$
But I can't conclude from here.



I sincerely thank you for your help.






metric-spaces norm normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Thank you for your correction.
    $endgroup$
    – Furdzik
    Jan 5 at 9:58










  • $begingroup$
    @PaulFrost it does not satisfy $f(0)=0$.
    $endgroup$
    – Lorenzo Quarisa
    Jan 5 at 11:07
















5












$begingroup$


I can't show that the two norms defined as
$$||f||_{infty}=sup_{xin[0,1]}|f(x)+ f'(x)|$$
and
$$N(f)=sup_{xin[0,1]}|f(x)| + sup_{xin[0,1]}|f'(x)|$$
are equivalent in $E={finmathcal{C}^1([0,1]) text{ s.t. } f(0)=0}$.



A first inequality in one sense is trivial.



For the other inequality, I can write:



$$
f(x)=f(0)+int_0^x f'(t) dt
$$
Which implies
$$
sup_{xin[0,1]}|f(x)| leq sup_{xin[0,1]}|f'(x)| .
$$
And then
$$
N(f)leq 2 sup_{xin[0,1]}|f'(x)| .
$$
But I can't conclude from here.



I sincerely thank you for your help.






metric-spaces norm normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Thank you for your correction.
    $endgroup$
    – Furdzik
    Jan 5 at 9:58










  • $begingroup$
    @PaulFrost it does not satisfy $f(0)=0$.
    $endgroup$
    – Lorenzo Quarisa
    Jan 5 at 11:07














5












5








5


1



$begingroup$


I can't show that the two norms defined as
$$||f||_{infty}=sup_{xin[0,1]}|f(x)+ f'(x)|$$
and
$$N(f)=sup_{xin[0,1]}|f(x)| + sup_{xin[0,1]}|f'(x)|$$
are equivalent in $E={finmathcal{C}^1([0,1]) text{ s.t. } f(0)=0}$.



A first inequality in one sense is trivial.



For the other inequality, I can write:



$$
f(x)=f(0)+int_0^x f'(t) dt
$$
Which implies
$$
sup_{xin[0,1]}|f(x)| leq sup_{xin[0,1]}|f'(x)| .
$$
And then
$$
N(f)leq 2 sup_{xin[0,1]}|f'(x)| .
$$
But I can't conclude from here.



I sincerely thank you for your help.






metric-spaces norm normed-spaces







share|cite|improve this question











$endgroup$




I can't show that the two norms defined as
$$||f||_{infty}=sup_{xin[0,1]}|f(x)+ f'(x)|$$
and
$$N(f)=sup_{xin[0,1]}|f(x)| + sup_{xin[0,1]}|f'(x)|$$
are equivalent in $E={finmathcal{C}^1([0,1]) text{ s.t. } f(0)=0}$.



A first inequality in one sense is trivial.



For the other inequality, I can write:



$$
f(x)=f(0)+int_0^x f'(t) dt
$$
Which implies
$$
sup_{xin[0,1]}|f(x)| leq sup_{xin[0,1]}|f'(x)| .
$$
And then
$$
N(f)leq 2 sup_{xin[0,1]}|f'(x)| .
$$
But I can't conclude from here.



I sincerely thank you for your help.






metric-spaces norm normed-spaces




metric-spaces norm normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 12:33









6005

36.3k751125




36.3k751125










asked Jan 5 at 9:47









FurdzikFurdzik

386




386












  • $begingroup$
    Thank you for your correction.
    $endgroup$
    – Furdzik
    Jan 5 at 9:58










  • $begingroup$
    @PaulFrost it does not satisfy $f(0)=0$.
    $endgroup$
    – Lorenzo Quarisa
    Jan 5 at 11:07


















  • $begingroup$
    Thank you for your correction.
    $endgroup$
    – Furdzik
    Jan 5 at 9:58










  • $begingroup$
    @PaulFrost it does not satisfy $f(0)=0$.
    $endgroup$
    – Lorenzo Quarisa
    Jan 5 at 11:07
















$begingroup$
Thank you for your correction.
$endgroup$
– Furdzik
Jan 5 at 9:58




$begingroup$
Thank you for your correction.
$endgroup$
– Furdzik
Jan 5 at 9:58












$begingroup$
@PaulFrost it does not satisfy $f(0)=0$.
$endgroup$
– Lorenzo Quarisa
Jan 5 at 11:07




$begingroup$
@PaulFrost it does not satisfy $f(0)=0$.
$endgroup$
– Lorenzo Quarisa
Jan 5 at 11:07










2 Answers
2






active

oldest

votes


















2












$begingroup$

Two metrics on the same set are equivalent (i.e. generate the same topology ) iff they have the same set of convergent sequences, because the closure operator in a metric space is entirely and uniquely determined by the convergent sequences.



For brevity let $sup_{xin [0,1]}|g(x)|=|g|_S$ for any continuous $g:[0,1]to Bbb R.$



Idea: Consider that $f(x)+f'(x)=e^{-x}(e^xf(x))'.$



(1). If $lim_{nto infty}|f_n-f|_{infty}=0$:



Then $lim_{nto infty}|e^{-x}(e^x (f_n(x)-f(x))'|_S=0,$ which implies that $lim_{nto infty}|(e^x(f_n(x)-f(x))'|_S=0,$ which, since $f_n(0)=f(0)=0,$ implies that $$lim_{nto infty}|e^x(f_n(x)-f(x))|_S= lim_{nto infty}sup_{xin [0,1]} |int_0^x(e^t(f_n(t)-f(t))'dt,|=0$$ which implies that $$lim_{nto infty}|f_n-f|_S=0.$$ Now $|f'_n-f'|_Sleq |(f_n+f'_n)-(f+f')|_S+|f-f_n)|_S=|f_n-f|_{infty}+|f_n-f|_S, $ so we have $$lim_{nto infty}|f'_n-f'|_S=0.$$ Since $N(f_n-f)=|f_n-f|_S+|f'_n-f'|_S,$ therefore $lim_{nto infty}N(f_n-f)=0.$



(2). If $lim_{nto infty}N(f_n-f)=0$:



Since $N(f_n-f)geq |f_n-f|_{infty}, $ therefore $ lim_{nto infty}|f_n-f|_{infty}=0.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) le |f|_infty le MN(f)$ for all $f$.
    $endgroup$
    – Theo Bendit
    Jan 5 at 11:21






  • 1




    $begingroup$
    @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $|v_n|_{(1)}>n|v_n|_{(2)}$ for each $nin Bbb N.$ Let $w_n=frac {v_n}{|v_n|_{(1)}}.$ Then the sequence $(w_n)_{nin Bbb N}$ converges to $0$ with respect to $|cdot|_{(2)}$ but $|w_n|_{(1)}=1$.
    $endgroup$
    – DanielWainfleet
    Jan 5 at 12:19





















3












$begingroup$

As you have already shown, for $f in C^{1}[0,1]$ such that $f(0) = 0$,
$$
sup_{x in [0,1]} |f(x)| le sup_{x in [0,1]} |f'(x)|. tag{1}
$$



We can use this to show that $sup |f|$ is bounded by a constant times $sup |f + f'|$, as follows:
begin{align*}
sup_{x in [0,1]} |f(x)|
&le sup_{x in [0,1]} |e^x f(x)| \
&le sup_{x in [0,1]} left| frac{d}{dx} e^x f(x) right| qquadqquad text{(by (1), since $e^0 f(0) = 0$})\
&= sup_{x in [0,1]} |e^x f(x) + e^x f'(x)| \
&le e sup_{x in [0,1]} |f(x) + f'(x)|.
end{align*}



Now apply triangle inequality to $|f'| = |f' + f - f|$:
begin{align*}
sup |f'(x)| &= sup |f'(x) + f(x) - f(x)| \
&le sup |f(x) + f'(x)| + sup |f(x)| \
&le sup |f(x) + f'(x)| + e cdot sup |f(x) + f'(x)| \
&= (e + 1) sup |f(x) + f'(x)|.
end{align*}



Therefore, $sup |f(x)|$ and $sup |f'(x)|$ are both bounded above by a constant times $sup |f(x) + f'(x)|$. This shows that your two norms are equivalent.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Two metrics on the same set are equivalent (i.e. generate the same topology ) iff they have the same set of convergent sequences, because the closure operator in a metric space is entirely and uniquely determined by the convergent sequences.



    For brevity let $sup_{xin [0,1]}|g(x)|=|g|_S$ for any continuous $g:[0,1]to Bbb R.$



    Idea: Consider that $f(x)+f'(x)=e^{-x}(e^xf(x))'.$



    (1). If $lim_{nto infty}|f_n-f|_{infty}=0$:



    Then $lim_{nto infty}|e^{-x}(e^x (f_n(x)-f(x))'|_S=0,$ which implies that $lim_{nto infty}|(e^x(f_n(x)-f(x))'|_S=0,$ which, since $f_n(0)=f(0)=0,$ implies that $$lim_{nto infty}|e^x(f_n(x)-f(x))|_S= lim_{nto infty}sup_{xin [0,1]} |int_0^x(e^t(f_n(t)-f(t))'dt,|=0$$ which implies that $$lim_{nto infty}|f_n-f|_S=0.$$ Now $|f'_n-f'|_Sleq |(f_n+f'_n)-(f+f')|_S+|f-f_n)|_S=|f_n-f|_{infty}+|f_n-f|_S, $ so we have $$lim_{nto infty}|f'_n-f'|_S=0.$$ Since $N(f_n-f)=|f_n-f|_S+|f'_n-f'|_S,$ therefore $lim_{nto infty}N(f_n-f)=0.$



    (2). If $lim_{nto infty}N(f_n-f)=0$:



    Since $N(f_n-f)geq |f_n-f|_{infty}, $ therefore $ lim_{nto infty}|f_n-f|_{infty}=0.$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) le |f|_infty le MN(f)$ for all $f$.
      $endgroup$
      – Theo Bendit
      Jan 5 at 11:21






    • 1




      $begingroup$
      @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $|v_n|_{(1)}>n|v_n|_{(2)}$ for each $nin Bbb N.$ Let $w_n=frac {v_n}{|v_n|_{(1)}}.$ Then the sequence $(w_n)_{nin Bbb N}$ converges to $0$ with respect to $|cdot|_{(2)}$ but $|w_n|_{(1)}=1$.
      $endgroup$
      – DanielWainfleet
      Jan 5 at 12:19


















    2












    $begingroup$

    Two metrics on the same set are equivalent (i.e. generate the same topology ) iff they have the same set of convergent sequences, because the closure operator in a metric space is entirely and uniquely determined by the convergent sequences.



    For brevity let $sup_{xin [0,1]}|g(x)|=|g|_S$ for any continuous $g:[0,1]to Bbb R.$



    Idea: Consider that $f(x)+f'(x)=e^{-x}(e^xf(x))'.$



    (1). If $lim_{nto infty}|f_n-f|_{infty}=0$:



    Then $lim_{nto infty}|e^{-x}(e^x (f_n(x)-f(x))'|_S=0,$ which implies that $lim_{nto infty}|(e^x(f_n(x)-f(x))'|_S=0,$ which, since $f_n(0)=f(0)=0,$ implies that $$lim_{nto infty}|e^x(f_n(x)-f(x))|_S= lim_{nto infty}sup_{xin [0,1]} |int_0^x(e^t(f_n(t)-f(t))'dt,|=0$$ which implies that $$lim_{nto infty}|f_n-f|_S=0.$$ Now $|f'_n-f'|_Sleq |(f_n+f'_n)-(f+f')|_S+|f-f_n)|_S=|f_n-f|_{infty}+|f_n-f|_S, $ so we have $$lim_{nto infty}|f'_n-f'|_S=0.$$ Since $N(f_n-f)=|f_n-f|_S+|f'_n-f'|_S,$ therefore $lim_{nto infty}N(f_n-f)=0.$



    (2). If $lim_{nto infty}N(f_n-f)=0$:



    Since $N(f_n-f)geq |f_n-f|_{infty}, $ therefore $ lim_{nto infty}|f_n-f|_{infty}=0.$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) le |f|_infty le MN(f)$ for all $f$.
      $endgroup$
      – Theo Bendit
      Jan 5 at 11:21






    • 1




      $begingroup$
      @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $|v_n|_{(1)}>n|v_n|_{(2)}$ for each $nin Bbb N.$ Let $w_n=frac {v_n}{|v_n|_{(1)}}.$ Then the sequence $(w_n)_{nin Bbb N}$ converges to $0$ with respect to $|cdot|_{(2)}$ but $|w_n|_{(1)}=1$.
      $endgroup$
      – DanielWainfleet
      Jan 5 at 12:19
















    2












    2








    2





    $begingroup$

    Two metrics on the same set are equivalent (i.e. generate the same topology ) iff they have the same set of convergent sequences, because the closure operator in a metric space is entirely and uniquely determined by the convergent sequences.



    For brevity let $sup_{xin [0,1]}|g(x)|=|g|_S$ for any continuous $g:[0,1]to Bbb R.$



    Idea: Consider that $f(x)+f'(x)=e^{-x}(e^xf(x))'.$



    (1). If $lim_{nto infty}|f_n-f|_{infty}=0$:



    Then $lim_{nto infty}|e^{-x}(e^x (f_n(x)-f(x))'|_S=0,$ which implies that $lim_{nto infty}|(e^x(f_n(x)-f(x))'|_S=0,$ which, since $f_n(0)=f(0)=0,$ implies that $$lim_{nto infty}|e^x(f_n(x)-f(x))|_S= lim_{nto infty}sup_{xin [0,1]} |int_0^x(e^t(f_n(t)-f(t))'dt,|=0$$ which implies that $$lim_{nto infty}|f_n-f|_S=0.$$ Now $|f'_n-f'|_Sleq |(f_n+f'_n)-(f+f')|_S+|f-f_n)|_S=|f_n-f|_{infty}+|f_n-f|_S, $ so we have $$lim_{nto infty}|f'_n-f'|_S=0.$$ Since $N(f_n-f)=|f_n-f|_S+|f'_n-f'|_S,$ therefore $lim_{nto infty}N(f_n-f)=0.$



    (2). If $lim_{nto infty}N(f_n-f)=0$:



    Since $N(f_n-f)geq |f_n-f|_{infty}, $ therefore $ lim_{nto infty}|f_n-f|_{infty}=0.$






    share|cite|improve this answer











    $endgroup$



    Two metrics on the same set are equivalent (i.e. generate the same topology ) iff they have the same set of convergent sequences, because the closure operator in a metric space is entirely and uniquely determined by the convergent sequences.



    For brevity let $sup_{xin [0,1]}|g(x)|=|g|_S$ for any continuous $g:[0,1]to Bbb R.$



    Idea: Consider that $f(x)+f'(x)=e^{-x}(e^xf(x))'.$



    (1). If $lim_{nto infty}|f_n-f|_{infty}=0$:



    Then $lim_{nto infty}|e^{-x}(e^x (f_n(x)-f(x))'|_S=0,$ which implies that $lim_{nto infty}|(e^x(f_n(x)-f(x))'|_S=0,$ which, since $f_n(0)=f(0)=0,$ implies that $$lim_{nto infty}|e^x(f_n(x)-f(x))|_S= lim_{nto infty}sup_{xin [0,1]} |int_0^x(e^t(f_n(t)-f(t))'dt,|=0$$ which implies that $$lim_{nto infty}|f_n-f|_S=0.$$ Now $|f'_n-f'|_Sleq |(f_n+f'_n)-(f+f')|_S+|f-f_n)|_S=|f_n-f|_{infty}+|f_n-f|_S, $ so we have $$lim_{nto infty}|f'_n-f'|_S=0.$$ Since $N(f_n-f)=|f_n-f|_S+|f'_n-f'|_S,$ therefore $lim_{nto infty}N(f_n-f)=0.$



    (2). If $lim_{nto infty}N(f_n-f)=0$:



    Since $N(f_n-f)geq |f_n-f|_{infty}, $ therefore $ lim_{nto infty}|f_n-f|_{infty}=0.$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 5 at 12:07

























    answered Jan 5 at 11:16









    DanielWainfleetDanielWainfleet

    35.3k31648




    35.3k31648












    • $begingroup$
      Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) le |f|_infty le MN(f)$ for all $f$.
      $endgroup$
      – Theo Bendit
      Jan 5 at 11:21






    • 1




      $begingroup$
      @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $|v_n|_{(1)}>n|v_n|_{(2)}$ for each $nin Bbb N.$ Let $w_n=frac {v_n}{|v_n|_{(1)}}.$ Then the sequence $(w_n)_{nin Bbb N}$ converges to $0$ with respect to $|cdot|_{(2)}$ but $|w_n|_{(1)}=1$.
      $endgroup$
      – DanielWainfleet
      Jan 5 at 12:19




















    • $begingroup$
      Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) le |f|_infty le MN(f)$ for all $f$.
      $endgroup$
      – Theo Bendit
      Jan 5 at 11:21






    • 1




      $begingroup$
      @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $|v_n|_{(1)}>n|v_n|_{(2)}$ for each $nin Bbb N.$ Let $w_n=frac {v_n}{|v_n|_{(1)}}.$ Then the sequence $(w_n)_{nin Bbb N}$ converges to $0$ with respect to $|cdot|_{(2)}$ but $|w_n|_{(1)}=1$.
      $endgroup$
      – DanielWainfleet
      Jan 5 at 12:19


















    $begingroup$
    Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) le |f|_infty le MN(f)$ for all $f$.
    $endgroup$
    – Theo Bendit
    Jan 5 at 11:21




    $begingroup$
    Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) le |f|_infty le MN(f)$ for all $f$.
    $endgroup$
    – Theo Bendit
    Jan 5 at 11:21




    1




    1




    $begingroup$
    @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $|v_n|_{(1)}>n|v_n|_{(2)}$ for each $nin Bbb N.$ Let $w_n=frac {v_n}{|v_n|_{(1)}}.$ Then the sequence $(w_n)_{nin Bbb N}$ converges to $0$ with respect to $|cdot|_{(2)}$ but $|w_n|_{(1)}=1$.
    $endgroup$
    – DanielWainfleet
    Jan 5 at 12:19






    $begingroup$
    @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $|v_n|_{(1)}>n|v_n|_{(2)}$ for each $nin Bbb N.$ Let $w_n=frac {v_n}{|v_n|_{(1)}}.$ Then the sequence $(w_n)_{nin Bbb N}$ converges to $0$ with respect to $|cdot|_{(2)}$ but $|w_n|_{(1)}=1$.
    $endgroup$
    – DanielWainfleet
    Jan 5 at 12:19













    3












    $begingroup$

    As you have already shown, for $f in C^{1}[0,1]$ such that $f(0) = 0$,
    $$
    sup_{x in [0,1]} |f(x)| le sup_{x in [0,1]} |f'(x)|. tag{1}
    $$



    We can use this to show that $sup |f|$ is bounded by a constant times $sup |f + f'|$, as follows:
    begin{align*}
    sup_{x in [0,1]} |f(x)|
    &le sup_{x in [0,1]} |e^x f(x)| \
    &le sup_{x in [0,1]} left| frac{d}{dx} e^x f(x) right| qquadqquad text{(by (1), since $e^0 f(0) = 0$})\
    &= sup_{x in [0,1]} |e^x f(x) + e^x f'(x)| \
    &le e sup_{x in [0,1]} |f(x) + f'(x)|.
    end{align*}



    Now apply triangle inequality to $|f'| = |f' + f - f|$:
    begin{align*}
    sup |f'(x)| &= sup |f'(x) + f(x) - f(x)| \
    &le sup |f(x) + f'(x)| + sup |f(x)| \
    &le sup |f(x) + f'(x)| + e cdot sup |f(x) + f'(x)| \
    &= (e + 1) sup |f(x) + f'(x)|.
    end{align*}



    Therefore, $sup |f(x)|$ and $sup |f'(x)|$ are both bounded above by a constant times $sup |f(x) + f'(x)|$. This shows that your two norms are equivalent.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      As you have already shown, for $f in C^{1}[0,1]$ such that $f(0) = 0$,
      $$
      sup_{x in [0,1]} |f(x)| le sup_{x in [0,1]} |f'(x)|. tag{1}
      $$



      We can use this to show that $sup |f|$ is bounded by a constant times $sup |f + f'|$, as follows:
      begin{align*}
      sup_{x in [0,1]} |f(x)|
      &le sup_{x in [0,1]} |e^x f(x)| \
      &le sup_{x in [0,1]} left| frac{d}{dx} e^x f(x) right| qquadqquad text{(by (1), since $e^0 f(0) = 0$})\
      &= sup_{x in [0,1]} |e^x f(x) + e^x f'(x)| \
      &le e sup_{x in [0,1]} |f(x) + f'(x)|.
      end{align*}



      Now apply triangle inequality to $|f'| = |f' + f - f|$:
      begin{align*}
      sup |f'(x)| &= sup |f'(x) + f(x) - f(x)| \
      &le sup |f(x) + f'(x)| + sup |f(x)| \
      &le sup |f(x) + f'(x)| + e cdot sup |f(x) + f'(x)| \
      &= (e + 1) sup |f(x) + f'(x)|.
      end{align*}



      Therefore, $sup |f(x)|$ and $sup |f'(x)|$ are both bounded above by a constant times $sup |f(x) + f'(x)|$. This shows that your two norms are equivalent.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        As you have already shown, for $f in C^{1}[0,1]$ such that $f(0) = 0$,
        $$
        sup_{x in [0,1]} |f(x)| le sup_{x in [0,1]} |f'(x)|. tag{1}
        $$



        We can use this to show that $sup |f|$ is bounded by a constant times $sup |f + f'|$, as follows:
        begin{align*}
        sup_{x in [0,1]} |f(x)|
        &le sup_{x in [0,1]} |e^x f(x)| \
        &le sup_{x in [0,1]} left| frac{d}{dx} e^x f(x) right| qquadqquad text{(by (1), since $e^0 f(0) = 0$})\
        &= sup_{x in [0,1]} |e^x f(x) + e^x f'(x)| \
        &le e sup_{x in [0,1]} |f(x) + f'(x)|.
        end{align*}



        Now apply triangle inequality to $|f'| = |f' + f - f|$:
        begin{align*}
        sup |f'(x)| &= sup |f'(x) + f(x) - f(x)| \
        &le sup |f(x) + f'(x)| + sup |f(x)| \
        &le sup |f(x) + f'(x)| + e cdot sup |f(x) + f'(x)| \
        &= (e + 1) sup |f(x) + f'(x)|.
        end{align*}



        Therefore, $sup |f(x)|$ and $sup |f'(x)|$ are both bounded above by a constant times $sup |f(x) + f'(x)|$. This shows that your two norms are equivalent.






        share|cite|improve this answer









        $endgroup$



        As you have already shown, for $f in C^{1}[0,1]$ such that $f(0) = 0$,
        $$
        sup_{x in [0,1]} |f(x)| le sup_{x in [0,1]} |f'(x)|. tag{1}
        $$



        We can use this to show that $sup |f|$ is bounded by a constant times $sup |f + f'|$, as follows:
        begin{align*}
        sup_{x in [0,1]} |f(x)|
        &le sup_{x in [0,1]} |e^x f(x)| \
        &le sup_{x in [0,1]} left| frac{d}{dx} e^x f(x) right| qquadqquad text{(by (1), since $e^0 f(0) = 0$})\
        &= sup_{x in [0,1]} |e^x f(x) + e^x f'(x)| \
        &le e sup_{x in [0,1]} |f(x) + f'(x)|.
        end{align*}



        Now apply triangle inequality to $|f'| = |f' + f - f|$:
        begin{align*}
        sup |f'(x)| &= sup |f'(x) + f(x) - f(x)| \
        &le sup |f(x) + f'(x)| + sup |f(x)| \
        &le sup |f(x) + f'(x)| + e cdot sup |f(x) + f'(x)| \
        &= (e + 1) sup |f(x) + f'(x)|.
        end{align*}



        Therefore, $sup |f(x)|$ and $sup |f'(x)|$ are both bounded above by a constant times $sup |f(x) + f'(x)|$. This shows that your two norms are equivalent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 12:28









        60056005

        36.3k751125




        36.3k751125






























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