What functions have the following property?
up vote
2
down vote
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I'm looking for differentiable functions $f:Bbb RtoBbb R$ such that
$$left(int_0^1 |f(t)|dtright)^2> frac{f(0)^2+f(1)^2}2$$
I found $f(x)=k$, for some constant $k$, $f(x)=x$, $f(x)=x^2$ and $f(x)=e^x$ that hold the opposite, but I couldn't find any function with this property. What does a function need to have in order for the statement to hold?
functions
asked Dec 1 at 20:40
Garmekain
1,259719
add a comment |
up vote
2
down vote
favorite
I'm looking for differentiable functions $f:Bbb RtoBbb R$ such that
$$left(int_0^1 |f(t)|dtright)^2> frac{f(0)^2+f(1)^2}2$$
I found $f(x)=k$, for some constant $k$, $f(x)=x$, $f(x)=x^2$ and $f(x)=e^x$ that hold the opposite, but I couldn't find any function with this property. What does a function need to have in order for the statement to hold?
functions
asked Dec 1 at 20:40
Garmekain
1,259719
2
Any function with $f(0) = f(1) = 0$ and $f(x) not= 0$ for some $x in (0,1)$ will also do.
– MisterRiemann
Dec 1 at 20:42
@MisterRiemann Are they the only ones?
– Garmekain
Dec 1 at 20:46
1
Certainly not. Take e.g. $f(x) = pisin(pi x) + 1$. Then the LHS is $(2+1)^2=9$, and the RHS is $1$, but $f(0)=f(1)=1.$ There are a bunch of functions satisfying your property.
– MisterRiemann
Dec 1 at 20:51
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm looking for differentiable functions $f:Bbb RtoBbb R$ such that
$$left(int_0^1 |f(t)|dtright)^2> frac{f(0)^2+f(1)^2}2$$
I found $f(x)=k$, for some constant $k$, $f(x)=x$, $f(x)=x^2$ and $f(x)=e^x$ that hold the opposite, but I couldn't find any function with this property. What does a function need to have in order for the statement to hold?
functions
asked Dec 1 at 20:40
Garmekain
1,259719
I'm looking for differentiable functions $f:Bbb RtoBbb R$ such that
$$left(int_0^1 |f(t)|dtright)^2> frac{f(0)^2+f(1)^2}2$$
I found $f(x)=k$, for some constant $k$, $f(x)=x$, $f(x)=x^2$ and $f(x)=e^x$ that hold the opposite, but I couldn't find any function with this property. What does a function need to have in order for the statement to hold?
functions
functions
asked Dec 1 at 20:40
Garmekain
1,259719
asked Dec 1 at 20:40
Garmekain
1,259719
asked Dec 1 at 20:40
Garmekain
1,259719
asked Dec 1 at 20:40
Garmekain
1,259719
asked Dec 1 at 20:40
Garmekain
1,259719
1,259719
2
Any function with $f(0) = f(1) = 0$ and $f(x) not= 0$ for some $x in (0,1)$ will also do.
– MisterRiemann
Dec 1 at 20:42
@MisterRiemann Are they the only ones?
– Garmekain
Dec 1 at 20:46
1
Certainly not. Take e.g. $f(x) = pisin(pi x) + 1$. Then the LHS is $(2+1)^2=9$, and the RHS is $1$, but $f(0)=f(1)=1.$ There are a bunch of functions satisfying your property.
– MisterRiemann
Dec 1 at 20:51
add a comment |
2
Any function with $f(0) = f(1) = 0$ and $f(x) not= 0$ for some $x in (0,1)$ will also do.
– MisterRiemann
Dec 1 at 20:42
@MisterRiemann Are they the only ones?
– Garmekain
Dec 1 at 20:46
1
Certainly not. Take e.g. $f(x) = pisin(pi x) + 1$. Then the LHS is $(2+1)^2=9$, and the RHS is $1$, but $f(0)=f(1)=1.$ There are a bunch of functions satisfying your property.
– MisterRiemann
Dec 1 at 20:51
2
2
Any function with $f(0) = f(1) = 0$ and $f(x) not= 0$ for some $x in (0,1)$ will also do.
– MisterRiemann
Dec 1 at 20:42
Any function with $f(0) = f(1) = 0$ and $f(x) not= 0$ for some $x in (0,1)$ will also do.
– MisterRiemann
Dec 1 at 20:42
@MisterRiemann Are they the only ones?
– Garmekain
Dec 1 at 20:46
@MisterRiemann Are they the only ones?
– Garmekain
Dec 1 at 20:46
1
1
Certainly not. Take e.g. $f(x) = pisin(pi x) + 1$. Then the LHS is $(2+1)^2=9$, and the RHS is $1$, but $f(0)=f(1)=1.$ There are a bunch of functions satisfying your property.
– MisterRiemann
Dec 1 at 20:51
Certainly not. Take e.g. $f(x) = pisin(pi x) + 1$. Then the LHS is $(2+1)^2=9$, and the RHS is $1$, but $f(0)=f(1)=1.$ There are a bunch of functions satisfying your property.
– MisterRiemann
Dec 1 at 20:51
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Roughly speaking, any function with a large enough hump in the middle will satisfy this. The right side only considers the function values at the endpoints. If $f(frac 12)$ is very large and the function is continuous the left side can be made as large as you want. To make a specific one, just use a triangle
$$f(x)=begin {cases} 1+100x&0 le x lt frac 12\51-100(x-frac 12)&frac 12 le x le 1end {cases}$$
We have $f(0)=f(1)=1$ so the right side is $1$. The integral is $26$ so the left side is $676$
answered Dec 1 at 21:38
Ross Millikan
288k23195366
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Roughly speaking, any function with a large enough hump in the middle will satisfy this. The right side only considers the function values at the endpoints. If $f(frac 12)$ is very large and the function is continuous the left side can be made as large as you want. To make a specific one, just use a triangle
$$f(x)=begin {cases} 1+100x&0 le x lt frac 12\51-100(x-frac 12)&frac 12 le x le 1end {cases}$$
We have $f(0)=f(1)=1$ so the right side is $1$. The integral is $26$ so the left side is $676$
answered Dec 1 at 21:38
Ross Millikan
288k23195366
add a comment |
up vote
1
down vote
accepted
Roughly speaking, any function with a large enough hump in the middle will satisfy this. The right side only considers the function values at the endpoints. If $f(frac 12)$ is very large and the function is continuous the left side can be made as large as you want. To make a specific one, just use a triangle
$$f(x)=begin {cases} 1+100x&0 le x lt frac 12\51-100(x-frac 12)&frac 12 le x le 1end {cases}$$
We have $f(0)=f(1)=1$ so the right side is $1$. The integral is $26$ so the left side is $676$
answered Dec 1 at 21:38
Ross Millikan
288k23195366
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Roughly speaking, any function with a large enough hump in the middle will satisfy this. The right side only considers the function values at the endpoints. If $f(frac 12)$ is very large and the function is continuous the left side can be made as large as you want. To make a specific one, just use a triangle
$$f(x)=begin {cases} 1+100x&0 le x lt frac 12\51-100(x-frac 12)&frac 12 le x le 1end {cases}$$
We have $f(0)=f(1)=1$ so the right side is $1$. The integral is $26$ so the left side is $676$
answered Dec 1 at 21:38
Ross Millikan
288k23195366
Roughly speaking, any function with a large enough hump in the middle will satisfy this. The right side only considers the function values at the endpoints. If $f(frac 12)$ is very large and the function is continuous the left side can be made as large as you want. To make a specific one, just use a triangle
$$f(x)=begin {cases} 1+100x&0 le x lt frac 12\51-100(x-frac 12)&frac 12 le x le 1end {cases}$$
We have $f(0)=f(1)=1$ so the right side is $1$. The integral is $26$ so the left side is $676$
answered Dec 1 at 21:38
Ross Millikan
288k23195366
answered Dec 1 at 21:38
Ross Millikan
288k23195366
answered Dec 1 at 21:38
Ross Millikan
288k23195366
answered Dec 1 at 21:38
Ross Millikan
288k23195366
288k23195366
add a comment |
add a comment |
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Any function with $f(0) = f(1) = 0$ and $f(x) not= 0$ for some $x in (0,1)$ will also do.
– MisterRiemann
Dec 1 at 20:42
@MisterRiemann Are they the only ones?
– Garmekain
Dec 1 at 20:46
1
Certainly not. Take e.g. $f(x) = pisin(pi x) + 1$. Then the LHS is $(2+1)^2=9$, and the RHS is $1$, but $f(0)=f(1)=1.$ There are a bunch of functions satisfying your property.
– MisterRiemann
Dec 1 at 20:51