Show that the two norms in $E=left{fin mathcal{C}^1[0,1]:f(0)=0right}$ are equivalent?
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.$
$endgroup$
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.$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Jan 5 at 12:28
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$begingroup$
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