$f(a)=f(b)=f_{+}'(a)=f_{-}'(b)=0$ and $|f''(x)|le M$. Show that $|f(x)|le frac{M(b-a)^2}{16}$.
$begingroup$
Let $f:left[a, bright]tomathbb{R}$ be twice differentiable.
Suppose $f(a)=f(b)=f_{+}'(a)=f_{-}'(b)=0$ and $|f''(x)|le M$.
Show that $|f(x)|le frac{M(b-a)^2}{16}$.
My try:
From taylor expansion we have
$f(x)=frac{f''(xi _1)}{2}(x-a)^2$ and $f(x)=frac{f''(xi _2)}{2}(x-b)^2$.
Then $|f(x)|lefrac{M}{2}(x-a)^2$ and $|f(x)|lefrac{M}{2}(x-b)^2$
So $|f(x)|lemin{(frac{M}{2}(x-a)^2,frac{M}{2}(x-b)^2)}lefrac{M}{8}(b-a)^2$
But this conclusion is weaker than the aim.
And I thought through using
$$f(c)=frac{f''(xi )}{2}(c-a)(c-b)$$
I can say$$|f(x)|lefrac{M}{8}(b-a)^2$$
which only uses the condition that $f(a)=f(b)=0$ and is still weaker.
Any hints? Thank you in advance!
real-analysis
asked Dec 23 '18 at 3:20
ZeroZero
34810
$endgroup$
add a comment |
$begingroup$
Let $f:left[a, bright]tomathbb{R}$ be twice differentiable.
Suppose $f(a)=f(b)=f_{+}'(a)=f_{-}'(b)=0$ and $|f''(x)|le M$.
Show that $|f(x)|le frac{M(b-a)^2}{16}$.
My try:
From taylor expansion we have
$f(x)=frac{f''(xi _1)}{2}(x-a)^2$ and $f(x)=frac{f''(xi _2)}{2}(x-b)^2$.
Then $|f(x)|lefrac{M}{2}(x-a)^2$ and $|f(x)|lefrac{M}{2}(x-b)^2$
So $|f(x)|lemin{(frac{M}{2}(x-a)^2,frac{M}{2}(x-b)^2)}lefrac{M}{8}(b-a)^2$
But this conclusion is weaker than the aim.
And I thought through using
$$f(c)=frac{f''(xi )}{2}(c-a)(c-b)$$
I can say$$|f(x)|lefrac{M}{8}(b-a)^2$$
which only uses the condition that $f(a)=f(b)=0$ and is still weaker.
Any hints? Thank you in advance!
real-analysis
asked Dec 23 '18 at 3:20
ZeroZero
34810
$endgroup$
add a comment |
$begingroup$
Let $f:left[a, bright]tomathbb{R}$ be twice differentiable.
Suppose $f(a)=f(b)=f_{+}'(a)=f_{-}'(b)=0$ and $|f''(x)|le M$.
Show that $|f(x)|le frac{M(b-a)^2}{16}$.
My try:
From taylor expansion we have
$f(x)=frac{f''(xi _1)}{2}(x-a)^2$ and $f(x)=frac{f''(xi _2)}{2}(x-b)^2$.
Then $|f(x)|lefrac{M}{2}(x-a)^2$ and $|f(x)|lefrac{M}{2}(x-b)^2$
So $|f(x)|lemin{(frac{M}{2}(x-a)^2,frac{M}{2}(x-b)^2)}lefrac{M}{8}(b-a)^2$
But this conclusion is weaker than the aim.
And I thought through using
$$f(c)=frac{f''(xi )}{2}(c-a)(c-b)$$
I can say$$|f(x)|lefrac{M}{8}(b-a)^2$$
which only uses the condition that $f(a)=f(b)=0$ and is still weaker.
Any hints? Thank you in advance!
real-analysis
asked Dec 23 '18 at 3:20
ZeroZero
34810
$endgroup$
Let $f:left[a, bright]tomathbb{R}$ be twice differentiable.
Suppose $f(a)=f(b)=f_{+}'(a)=f_{-}'(b)=0$ and $|f''(x)|le M$.
Show that $|f(x)|le frac{M(b-a)^2}{16}$.
My try:
From taylor expansion we have
$f(x)=frac{f''(xi _1)}{2}(x-a)^2$ and $f(x)=frac{f''(xi _2)}{2}(x-b)^2$.
Then $|f(x)|lefrac{M}{2}(x-a)^2$ and $|f(x)|lefrac{M}{2}(x-b)^2$
So $|f(x)|lemin{(frac{M}{2}(x-a)^2,frac{M}{2}(x-b)^2)}lefrac{M}{8}(b-a)^2$
But this conclusion is weaker than the aim.
And I thought through using
$$f(c)=frac{f''(xi )}{2}(c-a)(c-b)$$
I can say$$|f(x)|lefrac{M}{8}(b-a)^2$$
which only uses the condition that $f(a)=f(b)=0$ and is still weaker.
Any hints? Thank you in advance!
real-analysis
real-analysis
asked Dec 23 '18 at 3:20
ZeroZero
34810
asked Dec 23 '18 at 3:20
ZeroZero
34810
edited Dec 23 '18 at 3:58
Zero
asked Dec 23 '18 at 3:20
ZeroZero
34810
asked Dec 23 '18 at 3:20
ZeroZero
34810
asked Dec 23 '18 at 3:20
ZeroZero
34810
34810
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $c in [a, b]$ be a point where $|f(x)|$ attains its maximum value on the interval. If $c=a$ or $c=b$ then $f$ is identically zero and we are done. Now assume $a < c < b$, then $f'(c) = 0$. We have to show that $|f(c)| le frac{M(b-a)^2}{16}$.
Case 1: If $a < c le frac{a+b}2$ then we can proceed as follows:
$f_{+}'(a) = f'(c) =0$ and $|f''(x)|le M$ implies
$$
|f'(x)| le M min(x-a, c-x ) quad text{for } a < x < c
$$
and therefore, using $f(a) = 0$,
$$
|f(c)| le int_a^c |f'(x)| , dt le M int_a^c min(x-a, c-x ) , dt \
= M left( int_a^{(a+c)/2} (x-a), dt + int_{(a+c)/2}^c (c-x) , dt right) \
= M frac{(c-a)^2}8 le M frac{(b-a)^2}{16} , .
$$
Case 2: If $frac{a+b}2 le c < b$ then we can estimate
$$
|f(c)| le int_c^b |f'(x)| , dt
$$
in the same way.
answered Dec 23 '18 at 6:35
Martin RMartin R
29k33458
$endgroup$
add a comment |
$begingroup$
I thought the method I mentioned can be modified.
Let $f(x_0)=displaystylemax_{[a,b]} f$, we aim to prove $f(x_0)lefrac{M}{16}(b-a)^2$.
Proof: According to the definition, we have $f'(x_0)=0$.
Without losing generality, we assume that $x_0in[a,frac{a+b}{2}]$
Because $|f'(x)|le M$, we have
$$f(x)le frac{1}{2}M(x-a)^2$$
In addition,
$$f(x)ge f(x_0)-frac{1}{2}M(x-x_0)^2$$
which leads to $$f(x_0)lefrac{M}{16}(b-a)^2$$
answered Dec 23 '18 at 6:36
ZeroZero
34810
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $c in [a, b]$ be a point where $|f(x)|$ attains its maximum value on the interval. If $c=a$ or $c=b$ then $f$ is identically zero and we are done. Now assume $a < c < b$, then $f'(c) = 0$. We have to show that $|f(c)| le frac{M(b-a)^2}{16}$.
Case 1: If $a < c le frac{a+b}2$ then we can proceed as follows:
$f_{+}'(a) = f'(c) =0$ and $|f''(x)|le M$ implies
$$
|f'(x)| le M min(x-a, c-x ) quad text{for } a < x < c
$$
and therefore, using $f(a) = 0$,
$$
|f(c)| le int_a^c |f'(x)| , dt le M int_a^c min(x-a, c-x ) , dt \
= M left( int_a^{(a+c)/2} (x-a), dt + int_{(a+c)/2}^c (c-x) , dt right) \
= M frac{(c-a)^2}8 le M frac{(b-a)^2}{16} , .
$$
Case 2: If $frac{a+b}2 le c < b$ then we can estimate
$$
|f(c)| le int_c^b |f'(x)| , dt
$$
in the same way.
answered Dec 23 '18 at 6:35
Martin RMartin R
29k33458
$endgroup$
add a comment |
$begingroup$
Let $c in [a, b]$ be a point where $|f(x)|$ attains its maximum value on the interval. If $c=a$ or $c=b$ then $f$ is identically zero and we are done. Now assume $a < c < b$, then $f'(c) = 0$. We have to show that $|f(c)| le frac{M(b-a)^2}{16}$.
Case 1: If $a < c le frac{a+b}2$ then we can proceed as follows:
$f_{+}'(a) = f'(c) =0$ and $|f''(x)|le M$ implies
$$
|f'(x)| le M min(x-a, c-x ) quad text{for } a < x < c
$$
and therefore, using $f(a) = 0$,
$$
|f(c)| le int_a^c |f'(x)| , dt le M int_a^c min(x-a, c-x ) , dt \
= M left( int_a^{(a+c)/2} (x-a), dt + int_{(a+c)/2}^c (c-x) , dt right) \
= M frac{(c-a)^2}8 le M frac{(b-a)^2}{16} , .
$$
Case 2: If $frac{a+b}2 le c < b$ then we can estimate
$$
|f(c)| le int_c^b |f'(x)| , dt
$$
in the same way.
answered Dec 23 '18 at 6:35
Martin RMartin R
29k33458
$endgroup$
add a comment |
$begingroup$
Let $c in [a, b]$ be a point where $|f(x)|$ attains its maximum value on the interval. If $c=a$ or $c=b$ then $f$ is identically zero and we are done. Now assume $a < c < b$, then $f'(c) = 0$. We have to show that $|f(c)| le frac{M(b-a)^2}{16}$.
Case 1: If $a < c le frac{a+b}2$ then we can proceed as follows:
$f_{+}'(a) = f'(c) =0$ and $|f''(x)|le M$ implies
$$
|f'(x)| le M min(x-a, c-x ) quad text{for } a < x < c
$$
and therefore, using $f(a) = 0$,
$$
|f(c)| le int_a^c |f'(x)| , dt le M int_a^c min(x-a, c-x ) , dt \
= M left( int_a^{(a+c)/2} (x-a), dt + int_{(a+c)/2}^c (c-x) , dt right) \
= M frac{(c-a)^2}8 le M frac{(b-a)^2}{16} , .
$$
Case 2: If $frac{a+b}2 le c < b$ then we can estimate
$$
|f(c)| le int_c^b |f'(x)| , dt
$$
in the same way.
answered Dec 23 '18 at 6:35
Martin RMartin R
29k33458
$endgroup$
Let $c in [a, b]$ be a point where $|f(x)|$ attains its maximum value on the interval. If $c=a$ or $c=b$ then $f$ is identically zero and we are done. Now assume $a < c < b$, then $f'(c) = 0$. We have to show that $|f(c)| le frac{M(b-a)^2}{16}$.
Case 1: If $a < c le frac{a+b}2$ then we can proceed as follows:
$f_{+}'(a) = f'(c) =0$ and $|f''(x)|le M$ implies
$$
|f'(x)| le M min(x-a, c-x ) quad text{for } a < x < c
$$
and therefore, using $f(a) = 0$,
$$
|f(c)| le int_a^c |f'(x)| , dt le M int_a^c min(x-a, c-x ) , dt \
= M left( int_a^{(a+c)/2} (x-a), dt + int_{(a+c)/2}^c (c-x) , dt right) \
= M frac{(c-a)^2}8 le M frac{(b-a)^2}{16} , .
$$
Case 2: If $frac{a+b}2 le c < b$ then we can estimate
$$
|f(c)| le int_c^b |f'(x)| , dt
$$
in the same way.
answered Dec 23 '18 at 6:35
Martin RMartin R
29k33458
edited Dec 23 '18 at 6:46
answered Dec 23 '18 at 6:35
Martin RMartin R
29k33458
answered Dec 23 '18 at 6:35
Martin RMartin R
29k33458
answered Dec 23 '18 at 6:35
Martin RMartin R
29k33458
29k33458
add a comment |
add a comment |
$begingroup$
I thought the method I mentioned can be modified.
Let $f(x_0)=displaystylemax_{[a,b]} f$, we aim to prove $f(x_0)lefrac{M}{16}(b-a)^2$.
Proof: According to the definition, we have $f'(x_0)=0$.
Without losing generality, we assume that $x_0in[a,frac{a+b}{2}]$
Because $|f'(x)|le M$, we have
$$f(x)le frac{1}{2}M(x-a)^2$$
In addition,
$$f(x)ge f(x_0)-frac{1}{2}M(x-x_0)^2$$
which leads to $$f(x_0)lefrac{M}{16}(b-a)^2$$
answered Dec 23 '18 at 6:36
ZeroZero
34810
$endgroup$
add a comment |
$begingroup$
I thought the method I mentioned can be modified.
Let $f(x_0)=displaystylemax_{[a,b]} f$, we aim to prove $f(x_0)lefrac{M}{16}(b-a)^2$.
Proof: According to the definition, we have $f'(x_0)=0$.
Without losing generality, we assume that $x_0in[a,frac{a+b}{2}]$
Because $|f'(x)|le M$, we have
$$f(x)le frac{1}{2}M(x-a)^2$$
In addition,
$$f(x)ge f(x_0)-frac{1}{2}M(x-x_0)^2$$
which leads to $$f(x_0)lefrac{M}{16}(b-a)^2$$
answered Dec 23 '18 at 6:36
ZeroZero
34810
$endgroup$
add a comment |
$begingroup$
I thought the method I mentioned can be modified.
Let $f(x_0)=displaystylemax_{[a,b]} f$, we aim to prove $f(x_0)lefrac{M}{16}(b-a)^2$.
Proof: According to the definition, we have $f'(x_0)=0$.
Without losing generality, we assume that $x_0in[a,frac{a+b}{2}]$
Because $|f'(x)|le M$, we have
$$f(x)le frac{1}{2}M(x-a)^2$$
In addition,
$$f(x)ge f(x_0)-frac{1}{2}M(x-x_0)^2$$
which leads to $$f(x_0)lefrac{M}{16}(b-a)^2$$
answered Dec 23 '18 at 6:36
ZeroZero
34810
$endgroup$
I thought the method I mentioned can be modified.
Let $f(x_0)=displaystylemax_{[a,b]} f$, we aim to prove $f(x_0)lefrac{M}{16}(b-a)^2$.
Proof: According to the definition, we have $f'(x_0)=0$.
Without losing generality, we assume that $x_0in[a,frac{a+b}{2}]$
Because $|f'(x)|le M$, we have
$$f(x)le frac{1}{2}M(x-a)^2$$
In addition,
$$f(x)ge f(x_0)-frac{1}{2}M(x-x_0)^2$$
which leads to $$f(x_0)lefrac{M}{16}(b-a)^2$$
answered Dec 23 '18 at 6:36
ZeroZero
34810
edited Dec 23 '18 at 14:51
answered Dec 23 '18 at 6:36
ZeroZero
34810
answered Dec 23 '18 at 6:36
ZeroZero
34810
answered Dec 23 '18 at 6:36
ZeroZero
34810
34810
add a comment |
add a comment |
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