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Convex function with modulus

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1

$begingroup$

Find the values of the real parameter $a$ such that the function
$f:[0,1] to R, f(x)=x^2-|x-a|$ is a convex function.
It is clear that if $a=0$ or $a=1$ the function will be the restriction of a quadratic function, thus convex.

And intutively f is not convex when $ a in (0,1)$ because in that case the function would have "a concavity" around the point $a$. but i would be interested in a rigorous approach in this case.

functions convex-analysis

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asked yesterday

amarius8312amarius8312

1897

$endgroup$

add a comment | 

1

$begingroup$

Find the values of the real parameter $a$ such that the function
$f:[0,1] to R, f(x)=x^2-|x-a|$ is a convex function.
It is clear that if $a=0$ or $a=1$ the function will be the restriction of a quadratic function, thus convex.

And intutively f is not convex when $ a in (0,1)$ because in that case the function would have "a concavity" around the point $a$. but i would be interested in a rigorous approach in this case.

functions convex-analysis

share|cite|improve this question

asked yesterday

amarius8312amarius8312

1897

$endgroup$

add a comment | 

$begingroup$

Find the values of the real parameter $a$ such that the function
$f:[0,1] to R, f(x)=x^2-|x-a|$ is a convex function.
It is clear that if $a=0$ or $a=1$ the function will be the restriction of a quadratic function, thus convex.

And intutively f is not convex when $ a in (0,1)$ because in that case the function would have "a concavity" around the point $a$. but i would be interested in a rigorous approach in this case.

functions convex-analysis

share|cite|improve this question

asked yesterday

amarius8312amarius8312

1897

$endgroup$

share|cite|improve this question

asked yesterday

amarius8312amarius8312

1897

share|cite|improve this question

asked yesterday

amarius8312amarius8312

1897

asked yesterday

amarius8312amarius8312

1897

asked yesterday

amarius8312amarius8312

1897

1 Answer
1

active

oldest

votes

2

$begingroup$

You said:

It is clear that if $a=0$ or $a=1$ the function will be the restriction of a quadratic function, thus convex.

That is correct, and holds even for $a le 0$ and for $a ge 1$.

Then you said:

And intutively f is not convex when $ a in (0,1)$ because in that case the function would have "a concavity" around the point $a$.

For a rigorous approach we can compute
$$
frac 12 bigl( f(a-h) + f(a+h)bigr) - f(a)
$$
which should be $ge 0$ for a convex function. But if $0 < a < 1$ and $0 < h < min(a, 1-a)$ (so that all terms are defined) this expression evaluates to
$$
h^2-2h = (1-h)^2 - 1 < 0 , ,
$$
so that $f$ is not convex.

Alternatively: If $f$ were convex then
$$
f(x) le (1-x)f(0) + x f(1) = (1-x)(-a) + xa = -a + 2ax
$$
for all $x in [0, 1]$, this is violated at $x=a$ with $f(a) = a^2$.

Yet another approach for the case $0 < a < 1$ would be to observe that
$$
f'(x) = begin{cases}
2x + 1 & text{for } 0 le x < a \
2x - 1 & text{for } a < x le 1 \
end{cases}
$$
which also demonstrates that $f$ is not convex on $[0, 1]$, because
a convex function has a right (and left) derivative at every point, and the right (and left) derivative is monotonically increasing.

share|cite|improve this answer

edited yesterday

answered yesterday

Martin RMartin R

29.7k33558

$endgroup$

add a comment | 

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2

$begingroup$

You said:

It is clear that if $a=0$ or $a=1$ the function will be the restriction of a quadratic function, thus convex.

That is correct, and holds even for $a le 0$ and for $a ge 1$.

Then you said:

And intutively f is not convex when $ a in (0,1)$ because in that case the function would have "a concavity" around the point $a$.

For a rigorous approach we can compute
$$
frac 12 bigl( f(a-h) + f(a+h)bigr) - f(a)
$$
which should be $ge 0$ for a convex function. But if $0 < a < 1$ and $0 < h < min(a, 1-a)$ (so that all terms are defined) this expression evaluates to
$$
h^2-2h = (1-h)^2 - 1 < 0 , ,
$$
so that $f$ is not convex.

Alternatively: If $f$ were convex then
$$
f(x) le (1-x)f(0) + x f(1) = (1-x)(-a) + xa = -a + 2ax
$$
for all $x in [0, 1]$, this is violated at $x=a$ with $f(a) = a^2$.

Yet another approach for the case $0 < a < 1$ would be to observe that
$$
f'(x) = begin{cases}
2x + 1 & text{for } 0 le x < a \
2x - 1 & text{for } a < x le 1 \
end{cases}
$$
which also demonstrates that $f$ is not convex on $[0, 1]$, because
a convex function has a right (and left) derivative at every point, and the right (and left) derivative is monotonically increasing.

share|cite|improve this answer

edited yesterday

answered yesterday

Martin RMartin R

29.7k33558

$endgroup$

add a comment | 

2

$begingroup$

You said:

It is clear that if $a=0$ or $a=1$ the function will be the restriction of a quadratic function, thus convex.

That is correct, and holds even for $a le 0$ and for $a ge 1$.

Then you said:

And intutively f is not convex when $ a in (0,1)$ because in that case the function would have "a concavity" around the point $a$.

For a rigorous approach we can compute
$$
frac 12 bigl( f(a-h) + f(a+h)bigr) - f(a)
$$
which should be $ge 0$ for a convex function. But if $0 < a < 1$ and $0 < h < min(a, 1-a)$ (so that all terms are defined) this expression evaluates to
$$
h^2-2h = (1-h)^2 - 1 < 0 , ,
$$
so that $f$ is not convex.

Alternatively: If $f$ were convex then
$$
f(x) le (1-x)f(0) + x f(1) = (1-x)(-a) + xa = -a + 2ax
$$
for all $x in [0, 1]$, this is violated at $x=a$ with $f(a) = a^2$.

Yet another approach for the case $0 < a < 1$ would be to observe that
$$
f'(x) = begin{cases}
2x + 1 & text{for } 0 le x < a \
2x - 1 & text{for } a < x le 1 \
end{cases}
$$
which also demonstrates that $f$ is not convex on $[0, 1]$, because
a convex function has a right (and left) derivative at every point, and the right (and left) derivative is monotonically increasing.

share|cite|improve this answer

edited yesterday

answered yesterday

Martin RMartin R

29.7k33558

$endgroup$

add a comment | 

$begingroup$

You said:

It is clear that if $a=0$ or $a=1$ the function will be the restriction of a quadratic function, thus convex.

That is correct, and holds even for $a le 0$ and for $a ge 1$.

Then you said:

And intutively f is not convex when $ a in (0,1)$ because in that case the function would have "a concavity" around the point $a$.

For a rigorous approach we can compute
$$
frac 12 bigl( f(a-h) + f(a+h)bigr) - f(a)
$$
which should be $ge 0$ for a convex function. But if $0 < a < 1$ and $0 < h < min(a, 1-a)$ (so that all terms are defined) this expression evaluates to
$$
h^2-2h = (1-h)^2 - 1 < 0 , ,
$$
so that $f$ is not convex.

Alternatively: If $f$ were convex then
$$
f(x) le (1-x)f(0) + x f(1) = (1-x)(-a) + xa = -a + 2ax
$$
for all $x in [0, 1]$, this is violated at $x=a$ with $f(a) = a^2$.

Yet another approach for the case $0 < a < 1$ would be to observe that
$$
f'(x) = begin{cases}
2x + 1 & text{for } 0 le x < a \
2x - 1 & text{for } a < x le 1 \
end{cases}
$$
which also demonstrates that $f$ is not convex on $[0, 1]$, because
a convex function has a right (and left) derivative at every point, and the right (and left) derivative is monotonically increasing.

share|cite|improve this answer

edited yesterday

answered yesterday

Martin RMartin R

29.7k33558

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Martin RMartin R

29.7k33558

answered yesterday

Martin RMartin R

29.7k33558

answered yesterday

Martin RMartin R

29.7k33558