Finding largest value for which $T$ maps $X$ into $X$
$begingroup$
Let $X_l = {f in C([0,l], mathbf{R}), 0 leq f(X) leq 2, forall x in [0,l]}$ and let $T: X rightarrow C([0,l], mathbf{R})$ be defined by $(T(f))(x) = int_0^{x} f(t)dt$
The question is to find the largest number $l_0$ such that $T$ maps $X_{l_0}$ into $X_{l_0}$. My first guess is that $l_0 = 1$ but I'm not sure how to go about showing that.
real-analysis
edited Dec 17 '18 at 4:01
Tianlalu
3,08421138
asked Dec 17 '18 at 3:25
hopelessundergradhopelessundergrad
292
$endgroup$
add a comment |
$begingroup$
Let $X_l = {f in C([0,l], mathbf{R}), 0 leq f(X) leq 2, forall x in [0,l]}$ and let $T: X rightarrow C([0,l], mathbf{R})$ be defined by $(T(f))(x) = int_0^{x} f(t)dt$
The question is to find the largest number $l_0$ such that $T$ maps $X_{l_0}$ into $X_{l_0}$. My first guess is that $l_0 = 1$ but I'm not sure how to go about showing that.
real-analysis
edited Dec 17 '18 at 4:01
Tianlalu
3,08421138
asked Dec 17 '18 at 3:25
hopelessundergradhopelessundergrad
292
$endgroup$
add a comment |
$begingroup$
Let $X_l = {f in C([0,l], mathbf{R}), 0 leq f(X) leq 2, forall x in [0,l]}$ and let $T: X rightarrow C([0,l], mathbf{R})$ be defined by $(T(f))(x) = int_0^{x} f(t)dt$
The question is to find the largest number $l_0$ such that $T$ maps $X_{l_0}$ into $X_{l_0}$. My first guess is that $l_0 = 1$ but I'm not sure how to go about showing that.
real-analysis
edited Dec 17 '18 at 4:01
Tianlalu
3,08421138
asked Dec 17 '18 at 3:25
hopelessundergradhopelessundergrad
292
$endgroup$
Let $X_l = {f in C([0,l], mathbf{R}), 0 leq f(X) leq 2, forall x in [0,l]}$ and let $T: X rightarrow C([0,l], mathbf{R})$ be defined by $(T(f))(x) = int_0^{x} f(t)dt$
The question is to find the largest number $l_0$ such that $T$ maps $X_{l_0}$ into $X_{l_0}$. My first guess is that $l_0 = 1$ but I'm not sure how to go about showing that.
real-analysis
real-analysis
edited Dec 17 '18 at 4:01
Tianlalu
3,08421138
asked Dec 17 '18 at 3:25
hopelessundergradhopelessundergrad
292
edited Dec 17 '18 at 4:01
Tianlalu
3,08421138
asked Dec 17 '18 at 3:25
hopelessundergradhopelessundergrad
292
edited Dec 17 '18 at 4:01
Tianlalu
3,08421138
edited Dec 17 '18 at 4:01
Tianlalu
3,08421138
edited Dec 17 '18 at 4:01
Tianlalu
3,08421138
3,08421138
asked Dec 17 '18 at 3:25
hopelessundergradhopelessundergrad
292
asked Dec 17 '18 at 3:25
hopelessundergradhopelessundergrad
292
asked Dec 17 '18 at 3:25
hopelessundergradhopelessundergrad
292
292
add a comment |
add a comment |
1 Answer
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$begingroup$
If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.
For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:
$$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$
In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.
Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:
$$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$
Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.
answered Dec 17 '18 at 4:30
M. NestorM. Nestor
791113
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.
For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:
$$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$
In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.
Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:
$$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$
Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.
answered Dec 17 '18 at 4:30
M. NestorM. Nestor
791113
$endgroup$
add a comment |
$begingroup$
If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.
For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:
$$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$
In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.
Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:
$$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$
Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.
answered Dec 17 '18 at 4:30
M. NestorM. Nestor
791113
$endgroup$
add a comment |
$begingroup$
If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.
For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:
$$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$
In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.
Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:
$$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$
Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.
answered Dec 17 '18 at 4:30
M. NestorM. Nestor
791113
$endgroup$
If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.
For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:
$$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$
In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.
Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:
$$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$
Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.
answered Dec 17 '18 at 4:30
M. NestorM. Nestor
791113
edited Dec 17 '18 at 5:33
answered Dec 17 '18 at 4:30
M. NestorM. Nestor
791113
answered Dec 17 '18 at 4:30
M. NestorM. Nestor
791113
answered Dec 17 '18 at 4:30
M. NestorM. Nestor
791113
791113
add a comment |
add a comment |
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