Pointwise bound on differentiable $f$ in terms of $|f|_2$ and $|f'|_2$.
I have a relatively simple analysis question that for some reason has me stuck.
Suppose that $f:mathbb Rtomathbb R$ satisfies the following conditions.
$f$ is differentiable
$|f|_2$ and $|f'|_2$ are finite
$f$ is compactly supported.
Then, a simple argument shows that $fleqsqrt{2|f|_2|f'|_2}$ pointwise:
$$|f(x)|^2=left|int_{-infty}^x(f(x)^2)'~dxright|leq2int_a^x |f(x)||f'(x)|~dxleq2|f|_2|f'|_2.tag{1}$$
Remark. Intuitively, this kind of makes sense: If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure. But then since $f$ is continuous it probably has to increase/decrease at a large rate next to its very large values, which is "ruled out" by $f'$ having small $L^2$-norm.
Question. Does a similar result hold when we consider functions on a bounded interval $f:[a,b]tomathbb R$?
Replicating the same argument here yields the bound
$$|f(x)^2-f(a)^2|=left|int_{a}^x(f(x)^2)'~dxright|leq 2|f|_2|f'|_2,$$
but then this is not quite the same as my first example: if we don't know how large $f(a)$ is then we don't have a pointwise bound of $f$. However, I don't see why the intuition I mention in the remak above doesn't apply here.
In short, is my intuition misguided, or am I generalizing the procedure in equation $(1)$ incorrectly?
calculus real-analysis functions
asked Nov 21 '18 at 16:47
user78270
2,158721
add a comment |
I have a relatively simple analysis question that for some reason has me stuck.
Suppose that $f:mathbb Rtomathbb R$ satisfies the following conditions.
$f$ is differentiable
$|f|_2$ and $|f'|_2$ are finite
$f$ is compactly supported.
Then, a simple argument shows that $fleqsqrt{2|f|_2|f'|_2}$ pointwise:
$$|f(x)|^2=left|int_{-infty}^x(f(x)^2)'~dxright|leq2int_a^x |f(x)||f'(x)|~dxleq2|f|_2|f'|_2.tag{1}$$
Remark. Intuitively, this kind of makes sense: If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure. But then since $f$ is continuous it probably has to increase/decrease at a large rate next to its very large values, which is "ruled out" by $f'$ having small $L^2$-norm.
Question. Does a similar result hold when we consider functions on a bounded interval $f:[a,b]tomathbb R$?
Replicating the same argument here yields the bound
$$|f(x)^2-f(a)^2|=left|int_{a}^x(f(x)^2)'~dxright|leq 2|f|_2|f'|_2,$$
but then this is not quite the same as my first example: if we don't know how large $f(a)$ is then we don't have a pointwise bound of $f$. However, I don't see why the intuition I mention in the remak above doesn't apply here.
In short, is my intuition misguided, or am I generalizing the procedure in equation $(1)$ incorrectly?
calculus real-analysis functions
asked Nov 21 '18 at 16:47
user78270
2,158721
add a comment |
I have a relatively simple analysis question that for some reason has me stuck.
Suppose that $f:mathbb Rtomathbb R$ satisfies the following conditions.
$f$ is differentiable
$|f|_2$ and $|f'|_2$ are finite
$f$ is compactly supported.
Then, a simple argument shows that $fleqsqrt{2|f|_2|f'|_2}$ pointwise:
$$|f(x)|^2=left|int_{-infty}^x(f(x)^2)'~dxright|leq2int_a^x |f(x)||f'(x)|~dxleq2|f|_2|f'|_2.tag{1}$$
Remark. Intuitively, this kind of makes sense: If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure. But then since $f$ is continuous it probably has to increase/decrease at a large rate next to its very large values, which is "ruled out" by $f'$ having small $L^2$-norm.
Question. Does a similar result hold when we consider functions on a bounded interval $f:[a,b]tomathbb R$?
Replicating the same argument here yields the bound
$$|f(x)^2-f(a)^2|=left|int_{a}^x(f(x)^2)'~dxright|leq 2|f|_2|f'|_2,$$
but then this is not quite the same as my first example: if we don't know how large $f(a)$ is then we don't have a pointwise bound of $f$. However, I don't see why the intuition I mention in the remak above doesn't apply here.
In short, is my intuition misguided, or am I generalizing the procedure in equation $(1)$ incorrectly?
calculus real-analysis functions
asked Nov 21 '18 at 16:47
user78270
2,158721
I have a relatively simple analysis question that for some reason has me stuck.
Suppose that $f:mathbb Rtomathbb R$ satisfies the following conditions.
$f$ is differentiable
$|f|_2$ and $|f'|_2$ are finite
$f$ is compactly supported.
Then, a simple argument shows that $fleqsqrt{2|f|_2|f'|_2}$ pointwise:
$$|f(x)|^2=left|int_{-infty}^x(f(x)^2)'~dxright|leq2int_a^x |f(x)||f'(x)|~dxleq2|f|_2|f'|_2.tag{1}$$
Remark. Intuitively, this kind of makes sense: If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure. But then since $f$ is continuous it probably has to increase/decrease at a large rate next to its very large values, which is "ruled out" by $f'$ having small $L^2$-norm.
Question. Does a similar result hold when we consider functions on a bounded interval $f:[a,b]tomathbb R$?
Replicating the same argument here yields the bound
$$|f(x)^2-f(a)^2|=left|int_{a}^x(f(x)^2)'~dxright|leq 2|f|_2|f'|_2,$$
but then this is not quite the same as my first example: if we don't know how large $f(a)$ is then we don't have a pointwise bound of $f$. However, I don't see why the intuition I mention in the remak above doesn't apply here.
In short, is my intuition misguided, or am I generalizing the procedure in equation $(1)$ incorrectly?
calculus real-analysis functions
calculus real-analysis functions
asked Nov 21 '18 at 16:47
user78270
2,158721
asked Nov 21 '18 at 16:47
user78270
2,158721
asked Nov 21 '18 at 16:47
user78270
2,158721
asked Nov 21 '18 at 16:47
user78270
2,158721
asked Nov 21 '18 at 16:47
user78270
2,158721
2,158721
add a comment |
add a comment |
1 Answer
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If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure
It depends on what you mean by 'very large'. If you remove the restriction of compact support, then you also consider the constant functions $f(x)=c$ for all $xin [a,b]$. If we increase the constant $c$, these can take arbitrarily large values over the whole $[a,b]$ even if the derivative is $0$ everywhere for any choice of $c$.
In particular, in the RHS we have $$2|f|_2|f'|_2=0$$ which means that the correct term in the LHS is indeed $|f(x)^2-f(a)^2|$ and not $|f(x)|^2$ by itself or anything 'similar'. Your inequality is sharp.
answered Nov 21 '18 at 18:59
Lorenzo Quarisa
3,166517
Makes perfect sense, thanks for clarifying my intuition.
– user78270
Nov 21 '18 at 19:29
add a comment |
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1 Answer
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If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure
It depends on what you mean by 'very large'. If you remove the restriction of compact support, then you also consider the constant functions $f(x)=c$ for all $xin [a,b]$. If we increase the constant $c$, these can take arbitrarily large values over the whole $[a,b]$ even if the derivative is $0$ everywhere for any choice of $c$.
In particular, in the RHS we have $$2|f|_2|f'|_2=0$$ which means that the correct term in the LHS is indeed $|f(x)^2-f(a)^2|$ and not $|f(x)|^2$ by itself or anything 'similar'. Your inequality is sharp.
answered Nov 21 '18 at 18:59
Lorenzo Quarisa
3,166517
Makes perfect sense, thanks for clarifying my intuition.
– user78270
Nov 21 '18 at 19:29
add a comment |
If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure
It depends on what you mean by 'very large'. If you remove the restriction of compact support, then you also consider the constant functions $f(x)=c$ for all $xin [a,b]$. If we increase the constant $c$, these can take arbitrarily large values over the whole $[a,b]$ even if the derivative is $0$ everywhere for any choice of $c$.
In particular, in the RHS we have $$2|f|_2|f'|_2=0$$ which means that the correct term in the LHS is indeed $|f(x)^2-f(a)^2|$ and not $|f(x)|^2$ by itself or anything 'similar'. Your inequality is sharp.
answered Nov 21 '18 at 18:59
Lorenzo Quarisa
3,166517
Makes perfect sense, thanks for clarifying my intuition.
– user78270
Nov 21 '18 at 19:29
add a comment |
If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure
It depends on what you mean by 'very large'. If you remove the restriction of compact support, then you also consider the constant functions $f(x)=c$ for all $xin [a,b]$. If we increase the constant $c$, these can take arbitrarily large values over the whole $[a,b]$ even if the derivative is $0$ everywhere for any choice of $c$.
In particular, in the RHS we have $$2|f|_2|f'|_2=0$$ which means that the correct term in the LHS is indeed $|f(x)^2-f(a)^2|$ and not $|f(x)|^2$ by itself or anything 'similar'. Your inequality is sharp.
answered Nov 21 '18 at 18:59
Lorenzo Quarisa
3,166517
If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure
It depends on what you mean by 'very large'. If you remove the restriction of compact support, then you also consider the constant functions $f(x)=c$ for all $xin [a,b]$. If we increase the constant $c$, these can take arbitrarily large values over the whole $[a,b]$ even if the derivative is $0$ everywhere for any choice of $c$.
In particular, in the RHS we have $$2|f|_2|f'|_2=0$$ which means that the correct term in the LHS is indeed $|f(x)^2-f(a)^2|$ and not $|f(x)|^2$ by itself or anything 'similar'. Your inequality is sharp.
answered Nov 21 '18 at 18:59
Lorenzo Quarisa
3,166517
answered Nov 21 '18 at 18:59
Lorenzo Quarisa
3,166517
answered Nov 21 '18 at 18:59
Lorenzo Quarisa
3,166517
answered Nov 21 '18 at 18:59
Lorenzo Quarisa
3,166517
3,166517
Makes perfect sense, thanks for clarifying my intuition.
– user78270
Nov 21 '18 at 19:29
add a comment |
Makes perfect sense, thanks for clarifying my intuition.
– user78270
Nov 21 '18 at 19:29
Makes perfect sense, thanks for clarifying my intuition.
– user78270
Nov 21 '18 at 19:29
Makes perfect sense, thanks for clarifying my intuition.
– user78270
Nov 21 '18 at 19:29
add a comment |
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