Continuity of a solution of the equation $f(x) - overline{f} = 0$.
$begingroup$
Consider a positive semi-definite function $f(x) in mathcal{L}^2left( mathbb{R} right)$ of class $textbf{C}^0$. I am interested in knowing the values of $x$ beyond which this function is uniformly below some value $overline{f}$.
$$ overline{x} triangleq min left{ x , Big vert , fleft( rright) leq overline{f}, ,, forall r > x right} $$
We will call $overline{x}$ to be the cut-off point of $f(x)$, and we will assume that $overline{x}$ exists.
Let $overline{x}_alpha$ represent the cut-off point of the function $alpha f(x)$, where $alpha$ is a positive real number, then I would like to prove that
$$ lim_{alpha to 1} overline{x}_alpha =overline{x} $$
Here too, we will assume that $overline{x}_alpha$ exists.
My tentative approach for the proof
In the case of an affine function $f(s) = mx + c$, then for any given $alpha$, we can find an exact value of $overline{x}_alpha$, which is
$$ overline{x}_alpha = left(frac{1-alpha}{alpha}right)frac{overline{f}}{m}$$
In the affine case, $lim_{alpha to 1} overline{x}_alpha = overline{x}$.
For other functions, we can find an approximate value for $overline{x}_alpha$ using an affine approximation the function $alpha f(x)$
$$ hat{overline{x}}_alpha = left(frac{1 - alpha}{alpha}right) frac{overline{f}}{f'(overline{x})} $$
In case $f'(x)$ does not exist, then we could replace it by $f'(overline{x}^-)$ or $f'(overline{x}^+)$ depending on whether $alpha < 1$ or $alpha > 1$.
Depending on the value of $alpha$ and the natuee of $f(x)$, we know that $hat{overline{x}}_alpha approx overline{x}$.
However, we know that $lim_{alpha to 1} hat{overline{x}}_alpha = overline{x}$. Does this also mean that $lim_{alpha to 1} overline{x}_alpha = overline(x)$? If so, how can we demonstrate this?
I think one way to do this could be to show that $alpha to 1$, implies that $hat{overline{x}}_alpha to overline{x}_alpha$. This would in turn imply that $overline{x}_alpha to overline{x}$.
real-analysis limits continuity
asked Jan 12 at 7:19
siva82kbsiva82kb
1408
$endgroup$
add a comment |
$begingroup$
Consider a positive semi-definite function $f(x) in mathcal{L}^2left( mathbb{R} right)$ of class $textbf{C}^0$. I am interested in knowing the values of $x$ beyond which this function is uniformly below some value $overline{f}$.
$$ overline{x} triangleq min left{ x , Big vert , fleft( rright) leq overline{f}, ,, forall r > x right} $$
We will call $overline{x}$ to be the cut-off point of $f(x)$, and we will assume that $overline{x}$ exists.
Let $overline{x}_alpha$ represent the cut-off point of the function $alpha f(x)$, where $alpha$ is a positive real number, then I would like to prove that
$$ lim_{alpha to 1} overline{x}_alpha =overline{x} $$
Here too, we will assume that $overline{x}_alpha$ exists.
My tentative approach for the proof
In the case of an affine function $f(s) = mx + c$, then for any given $alpha$, we can find an exact value of $overline{x}_alpha$, which is
$$ overline{x}_alpha = left(frac{1-alpha}{alpha}right)frac{overline{f}}{m}$$
In the affine case, $lim_{alpha to 1} overline{x}_alpha = overline{x}$.
For other functions, we can find an approximate value for $overline{x}_alpha$ using an affine approximation the function $alpha f(x)$
$$ hat{overline{x}}_alpha = left(frac{1 - alpha}{alpha}right) frac{overline{f}}{f'(overline{x})} $$
In case $f'(x)$ does not exist, then we could replace it by $f'(overline{x}^-)$ or $f'(overline{x}^+)$ depending on whether $alpha < 1$ or $alpha > 1$.
Depending on the value of $alpha$ and the natuee of $f(x)$, we know that $hat{overline{x}}_alpha approx overline{x}$.
However, we know that $lim_{alpha to 1} hat{overline{x}}_alpha = overline{x}$. Does this also mean that $lim_{alpha to 1} overline{x}_alpha = overline(x)$? If so, how can we demonstrate this?
I think one way to do this could be to show that $alpha to 1$, implies that $hat{overline{x}}_alpha to overline{x}_alpha$. This would in turn imply that $overline{x}_alpha to overline{x}$.
real-analysis limits continuity
asked Jan 12 at 7:19
siva82kbsiva82kb
1408
$endgroup$
add a comment |
$begingroup$
Consider a positive semi-definite function $f(x) in mathcal{L}^2left( mathbb{R} right)$ of class $textbf{C}^0$. I am interested in knowing the values of $x$ beyond which this function is uniformly below some value $overline{f}$.
$$ overline{x} triangleq min left{ x , Big vert , fleft( rright) leq overline{f}, ,, forall r > x right} $$
We will call $overline{x}$ to be the cut-off point of $f(x)$, and we will assume that $overline{x}$ exists.
Let $overline{x}_alpha$ represent the cut-off point of the function $alpha f(x)$, where $alpha$ is a positive real number, then I would like to prove that
$$ lim_{alpha to 1} overline{x}_alpha =overline{x} $$
Here too, we will assume that $overline{x}_alpha$ exists.
My tentative approach for the proof
In the case of an affine function $f(s) = mx + c$, then for any given $alpha$, we can find an exact value of $overline{x}_alpha$, which is
$$ overline{x}_alpha = left(frac{1-alpha}{alpha}right)frac{overline{f}}{m}$$
In the affine case, $lim_{alpha to 1} overline{x}_alpha = overline{x}$.
For other functions, we can find an approximate value for $overline{x}_alpha$ using an affine approximation the function $alpha f(x)$
$$ hat{overline{x}}_alpha = left(frac{1 - alpha}{alpha}right) frac{overline{f}}{f'(overline{x})} $$
In case $f'(x)$ does not exist, then we could replace it by $f'(overline{x}^-)$ or $f'(overline{x}^+)$ depending on whether $alpha < 1$ or $alpha > 1$.
Depending on the value of $alpha$ and the natuee of $f(x)$, we know that $hat{overline{x}}_alpha approx overline{x}$.
However, we know that $lim_{alpha to 1} hat{overline{x}}_alpha = overline{x}$. Does this also mean that $lim_{alpha to 1} overline{x}_alpha = overline(x)$? If so, how can we demonstrate this?
I think one way to do this could be to show that $alpha to 1$, implies that $hat{overline{x}}_alpha to overline{x}_alpha$. This would in turn imply that $overline{x}_alpha to overline{x}$.
real-analysis limits continuity
asked Jan 12 at 7:19
siva82kbsiva82kb
1408
$endgroup$
Consider a positive semi-definite function $f(x) in mathcal{L}^2left( mathbb{R} right)$ of class $textbf{C}^0$. I am interested in knowing the values of $x$ beyond which this function is uniformly below some value $overline{f}$.
$$ overline{x} triangleq min left{ x , Big vert , fleft( rright) leq overline{f}, ,, forall r > x right} $$
We will call $overline{x}$ to be the cut-off point of $f(x)$, and we will assume that $overline{x}$ exists.
Let $overline{x}_alpha$ represent the cut-off point of the function $alpha f(x)$, where $alpha$ is a positive real number, then I would like to prove that
$$ lim_{alpha to 1} overline{x}_alpha =overline{x} $$
Here too, we will assume that $overline{x}_alpha$ exists.
My tentative approach for the proof
In the case of an affine function $f(s) = mx + c$, then for any given $alpha$, we can find an exact value of $overline{x}_alpha$, which is
$$ overline{x}_alpha = left(frac{1-alpha}{alpha}right)frac{overline{f}}{m}$$
In the affine case, $lim_{alpha to 1} overline{x}_alpha = overline{x}$.
For other functions, we can find an approximate value for $overline{x}_alpha$ using an affine approximation the function $alpha f(x)$
$$ hat{overline{x}}_alpha = left(frac{1 - alpha}{alpha}right) frac{overline{f}}{f'(overline{x})} $$
In case $f'(x)$ does not exist, then we could replace it by $f'(overline{x}^-)$ or $f'(overline{x}^+)$ depending on whether $alpha < 1$ or $alpha > 1$.
Depending on the value of $alpha$ and the natuee of $f(x)$, we know that $hat{overline{x}}_alpha approx overline{x}$.
However, we know that $lim_{alpha to 1} hat{overline{x}}_alpha = overline{x}$. Does this also mean that $lim_{alpha to 1} overline{x}_alpha = overline(x)$? If so, how can we demonstrate this?
I think one way to do this could be to show that $alpha to 1$, implies that $hat{overline{x}}_alpha to overline{x}_alpha$. This would in turn imply that $overline{x}_alpha to overline{x}$.
real-analysis limits continuity
real-analysis limits continuity
asked Jan 12 at 7:19
siva82kbsiva82kb
1408
asked Jan 12 at 7:19
siva82kbsiva82kb
1408
asked Jan 12 at 7:19
siva82kbsiva82kb
1408
asked Jan 12 at 7:19
siva82kbsiva82kb
1408
asked Jan 12 at 7:19
siva82kbsiva82kb
1408
1408
add a comment |
add a comment |
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