$L^{infty}$ translation continuity
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Let $tau_h$ denote the the translation of $f$ by $h$, i.e $tau_h f(x) = f(x+h)$ . Let $f in L^infty$, then $f$ is not continuous under $h$, i.e $tau_h f$ does not converge to $f$ in the $infty$ norm as $h$ tends to $0$.
My proof: Consider the function given by $2xsin left(frac{1}{x}right) - cos (1/x)$ at $x neq 0$, and $0$ at $x = 0$, restricted to the interval $[-1,1]$. Then, we see that this function is bounded everywhere, but the translation operator is not continuous. Indeed, no matter how small we take $h$, we have at $x = 0$ that $|f_h - f|_infty = |f_h|_infty$ oscillates rapidly between $1$, $-1$, and thus we cannot have convergence under the supremum norm, as it will always bre greater than or equal to $2 -epsilon$ for some point.
real-analysis functional-analysis lp-spaces
asked Nov 28 at 1:09
rubikscube09
1,167717
add a comment |
up vote
1
down vote
favorite
Let $tau_h$ denote the the translation of $f$ by $h$, i.e $tau_h f(x) = f(x+h)$ . Let $f in L^infty$, then $f$ is not continuous under $h$, i.e $tau_h f$ does not converge to $f$ in the $infty$ norm as $h$ tends to $0$.
My proof: Consider the function given by $2xsin left(frac{1}{x}right) - cos (1/x)$ at $x neq 0$, and $0$ at $x = 0$, restricted to the interval $[-1,1]$. Then, we see that this function is bounded everywhere, but the translation operator is not continuous. Indeed, no matter how small we take $h$, we have at $x = 0$ that $|f_h - f|_infty = |f_h|_infty$ oscillates rapidly between $1$, $-1$, and thus we cannot have convergence under the supremum norm, as it will always bre greater than or equal to $2 -epsilon$ for some point.
real-analysis functional-analysis lp-spaces
asked Nov 28 at 1:09
rubikscube09
1,167717
1
Did you consider first $f = mathbf{1}_{[0, 1]}$ (the indicator function of the interval $[0, 1]$)? (I did not downvote the question.)
– Will M.
Nov 28 at 1:12
1
Wow. I feel silly
– rubikscube09
Nov 28 at 1:14
2
My instructor, many years past now, used to make us feel silly by giving us counterexamples with the zero or identity matrix (we would always complicate the results too much).
– Will M.
Nov 28 at 1:16
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $tau_h$ denote the the translation of $f$ by $h$, i.e $tau_h f(x) = f(x+h)$ . Let $f in L^infty$, then $f$ is not continuous under $h$, i.e $tau_h f$ does not converge to $f$ in the $infty$ norm as $h$ tends to $0$.
My proof: Consider the function given by $2xsin left(frac{1}{x}right) - cos (1/x)$ at $x neq 0$, and $0$ at $x = 0$, restricted to the interval $[-1,1]$. Then, we see that this function is bounded everywhere, but the translation operator is not continuous. Indeed, no matter how small we take $h$, we have at $x = 0$ that $|f_h - f|_infty = |f_h|_infty$ oscillates rapidly between $1$, $-1$, and thus we cannot have convergence under the supremum norm, as it will always bre greater than or equal to $2 -epsilon$ for some point.
real-analysis functional-analysis lp-spaces
asked Nov 28 at 1:09
rubikscube09
1,167717
Let $tau_h$ denote the the translation of $f$ by $h$, i.e $tau_h f(x) = f(x+h)$ . Let $f in L^infty$, then $f$ is not continuous under $h$, i.e $tau_h f$ does not converge to $f$ in the $infty$ norm as $h$ tends to $0$.
My proof: Consider the function given by $2xsin left(frac{1}{x}right) - cos (1/x)$ at $x neq 0$, and $0$ at $x = 0$, restricted to the interval $[-1,1]$. Then, we see that this function is bounded everywhere, but the translation operator is not continuous. Indeed, no matter how small we take $h$, we have at $x = 0$ that $|f_h - f|_infty = |f_h|_infty$ oscillates rapidly between $1$, $-1$, and thus we cannot have convergence under the supremum norm, as it will always bre greater than or equal to $2 -epsilon$ for some point.
real-analysis functional-analysis lp-spaces
real-analysis functional-analysis lp-spaces
asked Nov 28 at 1:09
rubikscube09
1,167717
asked Nov 28 at 1:09
rubikscube09
1,167717
asked Nov 28 at 1:09
rubikscube09
1,167717
asked Nov 28 at 1:09
rubikscube09
1,167717
asked Nov 28 at 1:09
rubikscube09
1,167717
1,167717
1
Did you consider first $f = mathbf{1}_{[0, 1]}$ (the indicator function of the interval $[0, 1]$)? (I did not downvote the question.)
– Will M.
Nov 28 at 1:12
1
Wow. I feel silly
– rubikscube09
Nov 28 at 1:14
2
My instructor, many years past now, used to make us feel silly by giving us counterexamples with the zero or identity matrix (we would always complicate the results too much).
– Will M.
Nov 28 at 1:16
add a comment |
1
Did you consider first $f = mathbf{1}_{[0, 1]}$ (the indicator function of the interval $[0, 1]$)? (I did not downvote the question.)
– Will M.
Nov 28 at 1:12
1
Wow. I feel silly
– rubikscube09
Nov 28 at 1:14
2
My instructor, many years past now, used to make us feel silly by giving us counterexamples with the zero or identity matrix (we would always complicate the results too much).
– Will M.
Nov 28 at 1:16
1
1
Did you consider first $f = mathbf{1}_{[0, 1]}$ (the indicator function of the interval $[0, 1]$)? (I did not downvote the question.)
– Will M.
Nov 28 at 1:12
Did you consider first $f = mathbf{1}_{[0, 1]}$ (the indicator function of the interval $[0, 1]$)? (I did not downvote the question.)
– Will M.
Nov 28 at 1:12
1
1
Wow. I feel silly
– rubikscube09
Nov 28 at 1:14
Wow. I feel silly
– rubikscube09
Nov 28 at 1:14
2
2
My instructor, many years past now, used to make us feel silly by giving us counterexamples with the zero or identity matrix (we would always complicate the results too much).
– Will M.
Nov 28 at 1:16
My instructor, many years past now, used to make us feel silly by giving us counterexamples with the zero or identity matrix (we would always complicate the results too much).
– Will M.
Nov 28 at 1:16
add a comment |
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1
Did you consider first $f = mathbf{1}_{[0, 1]}$ (the indicator function of the interval $[0, 1]$)? (I did not downvote the question.)
– Will M.
Nov 28 at 1:12
1
Wow. I feel silly
– rubikscube09
Nov 28 at 1:14
2
My instructor, many years past now, used to make us feel silly by giving us counterexamples with the zero or identity matrix (we would always complicate the results too much).
– Will M.
Nov 28 at 1:16