Relating the mean value theorem to the directional derivatives
$begingroup$
I am having some trouble understanding the process for proving the following statement:
If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.
In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 - x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:
$f(x_1 +v) -f(x_1) = f'(x_1 + theta v, v) = 0$
I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?
Thanks for your help.
calculus
asked Dec 6 '18 at 0:49
ChristianChristian
388
$endgroup$
add a comment |
$begingroup$
I am having some trouble understanding the process for proving the following statement:
If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.
In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 - x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:
$f(x_1 +v) -f(x_1) = f'(x_1 + theta v, v) = 0$
I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?
Thanks for your help.
calculus
asked Dec 6 '18 at 0:49
ChristianChristian
388
$endgroup$
$begingroup$
You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
$endgroup$
– famesyasd
Dec 6 '18 at 2:27
$begingroup$
What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
$endgroup$
– famesyasd
Dec 6 '18 at 2:31
$begingroup$
Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
$endgroup$
– famesyasd
Dec 6 '18 at 2:37
$begingroup$
I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
$endgroup$
– famesyasd
Dec 6 '18 at 2:38
$begingroup$
Thanks famesyasd - that is clear now.
$endgroup$
– Christian
Dec 6 '18 at 12:06
add a comment |
$begingroup$
I am having some trouble understanding the process for proving the following statement:
If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.
In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 - x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:
$f(x_1 +v) -f(x_1) = f'(x_1 + theta v, v) = 0$
I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?
Thanks for your help.
calculus
asked Dec 6 '18 at 0:49
ChristianChristian
388
$endgroup$
I am having some trouble understanding the process for proving the following statement:
If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.
In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 - x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:
$f(x_1 +v) -f(x_1) = f'(x_1 + theta v, v) = 0$
I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?
Thanks for your help.
calculus
calculus
asked Dec 6 '18 at 0:49
ChristianChristian
388
asked Dec 6 '18 at 0:49
ChristianChristian
388
asked Dec 6 '18 at 0:49
ChristianChristian
388
asked Dec 6 '18 at 0:49
ChristianChristian
388
asked Dec 6 '18 at 0:49
ChristianChristian
388
388
$begingroup$
You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
$endgroup$
– famesyasd
Dec 6 '18 at 2:27
$begingroup$
What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
$endgroup$
– famesyasd
Dec 6 '18 at 2:31
$begingroup$
Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
$endgroup$
– famesyasd
Dec 6 '18 at 2:37
$begingroup$
I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
$endgroup$
– famesyasd
Dec 6 '18 at 2:38
$begingroup$
Thanks famesyasd - that is clear now.
$endgroup$
– Christian
Dec 6 '18 at 12:06
add a comment |
$begingroup$
You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
$endgroup$
– famesyasd
Dec 6 '18 at 2:27
$begingroup$
What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
$endgroup$
– famesyasd
Dec 6 '18 at 2:31
$begingroup$
Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
$endgroup$
– famesyasd
Dec 6 '18 at 2:37
$begingroup$
I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
$endgroup$
– famesyasd
Dec 6 '18 at 2:38
$begingroup$
Thanks famesyasd - that is clear now.
$endgroup$
– Christian
Dec 6 '18 at 12:06
$begingroup$
You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
$endgroup$
– famesyasd
Dec 6 '18 at 2:27
$begingroup$
You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
$endgroup$
– famesyasd
Dec 6 '18 at 2:27
$begingroup$
What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
$endgroup$
– famesyasd
Dec 6 '18 at 2:31
$begingroup$
What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
$endgroup$
– famesyasd
Dec 6 '18 at 2:31
$begingroup$
Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
$endgroup$
– famesyasd
Dec 6 '18 at 2:37
$begingroup$
Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
$endgroup$
– famesyasd
Dec 6 '18 at 2:37
$begingroup$
I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
$endgroup$
– famesyasd
Dec 6 '18 at 2:38
$begingroup$
I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
$endgroup$
– famesyasd
Dec 6 '18 at 2:38
$begingroup$
Thanks famesyasd - that is clear now.
$endgroup$
– Christian
Dec 6 '18 at 12:06
$begingroup$
Thanks famesyasd - that is clear now.
$endgroup$
– Christian
Dec 6 '18 at 12:06
add a comment |
1 Answer
1
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oldest
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$newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
$$
lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
$$
and note that
$$
f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
$$
provided that the limit exists, we have
$$
lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
$$
i.e.
$$
F'(k) = f'(bm x_1 +kbm v , bm v).
$$
So by Lagrange's MVT,
$$
F(1) - F(0) = 1 cdot F'(k),
$$
which is just
$$
f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
$$
as shown in the OP.
To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.
answered Dec 6 '18 at 5:24
xbhxbh
6,3001522
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
$newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
$$
lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
$$
and note that
$$
f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
$$
provided that the limit exists, we have
$$
lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
$$
i.e.
$$
F'(k) = f'(bm x_1 +kbm v , bm v).
$$
So by Lagrange's MVT,
$$
F(1) - F(0) = 1 cdot F'(k),
$$
which is just
$$
f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
$$
as shown in the OP.
To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.
answered Dec 6 '18 at 5:24
xbhxbh
6,3001522
$endgroup$
add a comment |
$begingroup$
$newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
$$
lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
$$
and note that
$$
f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
$$
provided that the limit exists, we have
$$
lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
$$
i.e.
$$
F'(k) = f'(bm x_1 +kbm v , bm v).
$$
So by Lagrange's MVT,
$$
F(1) - F(0) = 1 cdot F'(k),
$$
which is just
$$
f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
$$
as shown in the OP.
To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.
answered Dec 6 '18 at 5:24
xbhxbh
6,3001522
$endgroup$
add a comment |
$begingroup$
$newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
$$
lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
$$
and note that
$$
f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
$$
provided that the limit exists, we have
$$
lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
$$
i.e.
$$
F'(k) = f'(bm x_1 +kbm v , bm v).
$$
So by Lagrange's MVT,
$$
F(1) - F(0) = 1 cdot F'(k),
$$
which is just
$$
f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
$$
as shown in the OP.
To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.
answered Dec 6 '18 at 5:24
xbhxbh
6,3001522
$endgroup$
$newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
$$
lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
$$
and note that
$$
f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
$$
provided that the limit exists, we have
$$
lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
$$
i.e.
$$
F'(k) = f'(bm x_1 +kbm v , bm v).
$$
So by Lagrange's MVT,
$$
F(1) - F(0) = 1 cdot F'(k),
$$
which is just
$$
f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
$$
as shown in the OP.
To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.
answered Dec 6 '18 at 5:24
xbhxbh
6,3001522
answered Dec 6 '18 at 5:24
xbhxbh
6,3001522
answered Dec 6 '18 at 5:24
xbhxbh
6,3001522
answered Dec 6 '18 at 5:24
xbhxbh
6,3001522
6,3001522
add a comment |
add a comment |
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$begingroup$
You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
$endgroup$
– famesyasd
Dec 6 '18 at 2:27
$begingroup$
What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
$endgroup$
– famesyasd
Dec 6 '18 at 2:31
$begingroup$
Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
$endgroup$
– famesyasd
Dec 6 '18 at 2:37
$begingroup$
I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
$endgroup$
– famesyasd
Dec 6 '18 at 2:38
$begingroup$
Thanks famesyasd - that is clear now.
$endgroup$
– Christian
Dec 6 '18 at 12:06