Determining whether $T$ is onto or not
Let $V=C[0,1]$, continuous real valued function on $[0,1]$
Consider the linear transformation $$T(f)=int_{0}^{x}f(t)dt$$
Is $T$ onto?
Attempts
What does onto mean? So for each $gin V$ we have some $fin V$ such that $T(f)=g$. In other words,
$$int_{0}^{x}f(t)dt=g(x)$$ Taking derivative on both sides(Using Leibniz Rule here) we get $f(x)=g'(x)$.
Take any function $g$ whose derivative does not exist in $[0,1]$ and we are done!.
Am I right??
Another approach would be to observe that constant function is not in the image of $T$.
If I am missing something please point out.
linear-algebra linear-transformations
edited Dec 27 '18 at 21:17
José Carlos Santos
151k22123224
asked Oct 16 '18 at 14:26
StammeringMathematician
2,2381322
add a comment |
Let $V=C[0,1]$, continuous real valued function on $[0,1]$
Consider the linear transformation $$T(f)=int_{0}^{x}f(t)dt$$
Is $T$ onto?
Attempts
What does onto mean? So for each $gin V$ we have some $fin V$ such that $T(f)=g$. In other words,
$$int_{0}^{x}f(t)dt=g(x)$$ Taking derivative on both sides(Using Leibniz Rule here) we get $f(x)=g'(x)$.
Take any function $g$ whose derivative does not exist in $[0,1]$ and we are done!.
Am I right??
Another approach would be to observe that constant function is not in the image of $T$.
If I am missing something please point out.
linear-algebra linear-transformations
edited Dec 27 '18 at 21:17
José Carlos Santos
151k22123224
asked Oct 16 '18 at 14:26
StammeringMathematician
2,2381322
add a comment |
Let $V=C[0,1]$, continuous real valued function on $[0,1]$
Consider the linear transformation $$T(f)=int_{0}^{x}f(t)dt$$
Is $T$ onto?
Attempts
What does onto mean? So for each $gin V$ we have some $fin V$ such that $T(f)=g$. In other words,
$$int_{0}^{x}f(t)dt=g(x)$$ Taking derivative on both sides(Using Leibniz Rule here) we get $f(x)=g'(x)$.
Take any function $g$ whose derivative does not exist in $[0,1]$ and we are done!.
Am I right??
Another approach would be to observe that constant function is not in the image of $T$.
If I am missing something please point out.
linear-algebra linear-transformations
edited Dec 27 '18 at 21:17
José Carlos Santos
151k22123224
asked Oct 16 '18 at 14:26
StammeringMathematician
2,2381322
Let $V=C[0,1]$, continuous real valued function on $[0,1]$
Consider the linear transformation $$T(f)=int_{0}^{x}f(t)dt$$
Is $T$ onto?
Attempts
What does onto mean? So for each $gin V$ we have some $fin V$ such that $T(f)=g$. In other words,
$$int_{0}^{x}f(t)dt=g(x)$$ Taking derivative on both sides(Using Leibniz Rule here) we get $f(x)=g'(x)$.
Take any function $g$ whose derivative does not exist in $[0,1]$ and we are done!.
Am I right??
Another approach would be to observe that constant function is not in the image of $T$.
If I am missing something please point out.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Dec 27 '18 at 21:17
José Carlos Santos
151k22123224
asked Oct 16 '18 at 14:26
StammeringMathematician
2,2381322
edited Dec 27 '18 at 21:17
José Carlos Santos
151k22123224
asked Oct 16 '18 at 14:26
StammeringMathematician
2,2381322
edited Dec 27 '18 at 21:17
José Carlos Santos
151k22123224
edited Dec 27 '18 at 21:17
José Carlos Santos
151k22123224
edited Dec 27 '18 at 21:17
José Carlos Santos
151k22123224
151k22123224
asked Oct 16 '18 at 14:26
StammeringMathematician
2,2381322
asked Oct 16 '18 at 14:26
StammeringMathematician
2,2381322
asked Oct 16 '18 at 14:26
StammeringMathematician
2,2381322
2,2381322
add a comment |
add a comment |
1 Answer
1
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oldest
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Yes, you are right.
Another way of proving that $T$ is not onto comes from noticing that you always have $T(f)(0)=0$. So, take a function $gin C[0,1]$ such that $g(0)neq0$ and you're done (again).
answered Oct 16 '18 at 14:29
José Carlos Santos
151k22123224
1
Wow!! This is simply great observation( at least at my level of mathematical maturity)
– StammeringMathematician
Oct 16 '18 at 14:34
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, you are right.
Another way of proving that $T$ is not onto comes from noticing that you always have $T(f)(0)=0$. So, take a function $gin C[0,1]$ such that $g(0)neq0$ and you're done (again).
answered Oct 16 '18 at 14:29
José Carlos Santos
151k22123224
1
Wow!! This is simply great observation( at least at my level of mathematical maturity)
– StammeringMathematician
Oct 16 '18 at 14:34
add a comment |
Yes, you are right.
Another way of proving that $T$ is not onto comes from noticing that you always have $T(f)(0)=0$. So, take a function $gin C[0,1]$ such that $g(0)neq0$ and you're done (again).
answered Oct 16 '18 at 14:29
José Carlos Santos
151k22123224
1
Wow!! This is simply great observation( at least at my level of mathematical maturity)
– StammeringMathematician
Oct 16 '18 at 14:34
add a comment |
Yes, you are right.
Another way of proving that $T$ is not onto comes from noticing that you always have $T(f)(0)=0$. So, take a function $gin C[0,1]$ such that $g(0)neq0$ and you're done (again).
answered Oct 16 '18 at 14:29
José Carlos Santos
151k22123224
Yes, you are right.
Another way of proving that $T$ is not onto comes from noticing that you always have $T(f)(0)=0$. So, take a function $gin C[0,1]$ such that $g(0)neq0$ and you're done (again).
answered Oct 16 '18 at 14:29
José Carlos Santos
151k22123224
edited Oct 16 '18 at 14:38
answered Oct 16 '18 at 14:29
José Carlos Santos
151k22123224
answered Oct 16 '18 at 14:29
José Carlos Santos
151k22123224
answered Oct 16 '18 at 14:29
José Carlos Santos
151k22123224
151k22123224
1
Wow!! This is simply great observation( at least at my level of mathematical maturity)
– StammeringMathematician
Oct 16 '18 at 14:34
add a comment |
1
Wow!! This is simply great observation( at least at my level of mathematical maturity)
– StammeringMathematician
Oct 16 '18 at 14:34
1
1
Wow!! This is simply great observation( at least at my level of mathematical maturity)
– StammeringMathematician
Oct 16 '18 at 14:34
Wow!! This is simply great observation( at least at my level of mathematical maturity)
– StammeringMathematician
Oct 16 '18 at 14:34
add a comment |
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