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0

$begingroup$

Let $f: mathbbR^m rightarrow mathbbR$. Show that $f(x) = x_i$ where $x_i$ is the $i^nd$ component of $x$ regarding the standard basis of $mathbbR^m$ is continuous in 0.

The definition says :

Let $A subset mathbbR^m$, $f: A rightarrow mathbbR^p$ and $a in A$. The function is continuous in a when for all $epsilon gt 0$, $exists delta gt 0$ such that for all $x in A$, $||x - a|| le delta$ and $|| f(x) - f(a) || le epsilon$.

I know $f(x) = (e_i|x)$.

By Cauchy-Schwarz inequality, we have : $|(e_i|x)| le ||e_i||.||x||$

We have $||x - 0|| = ||x|| le delta$.

So $||f(x) - f(0)|| le ||e_i||.||x|| - 0 = ||e_i||.||x||$

Now, for every $epsilon gt 0$ we can take $delta = fracepsilone_i$

So $||f(x) - f(0)|| le ||e_i||.||x|| le ||e_i||delta = ||e_i||fracepsilone_i = epsilon$.

I wonder if it's correct ?

real-analysis continuity vectors

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edited 16 hours ago

Mathieu Rousseau

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. But this is not a definition of continuity, I'm afraid. It should go as something like: is continuous at $a$ when for each $epsilon>0$ there is $delta>0$ such that for all $xin A$ such that $lVert x-arVert<delta$ there holds $lVert f(x)-f(a)rVert<epsilon$.
  $endgroup$
  – user539887
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry I made a mistake in the definition. I corrected it.
  $endgroup$
  – Mathieu Rousseau
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is O.K. When you use the standard Euclidean norm, then $lVert e_irVert=1$. But perhaps by $lVertcdotrVert$ you mean any arbitrary norm, then $lVert e_irVert$ need not be $1$. Incidentally, Cauchy-Schwarz inequality, not inegality.
  $endgroup$
  – user539887
  19 hours ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $f: mathbbR^m rightarrow mathbbR$. Show that $f(x) = x_i$ where $x_i$ is the $i^nd$ component of $x$ regarding the standard basis of $mathbbR^m$ is continuous in 0.

The definition says :

Let $A subset mathbbR^m$, $f: A rightarrow mathbbR^p$ and $a in A$. The function is continuous in a when for all $epsilon gt 0$, $exists delta gt 0$ such that for all $x in A$, $||x - a|| le delta$ and $|| f(x) - f(a) || le epsilon$.

I know $f(x) = (e_i|x)$.

By Cauchy-Schwarz inequality, we have : $|(e_i|x)| le ||e_i||.||x||$

We have $||x - 0|| = ||x|| le delta$.

So $||f(x) - f(0)|| le ||e_i||.||x|| - 0 = ||e_i||.||x||$

Now, for every $epsilon gt 0$ we can take $delta = fracepsilone_i$

So $||f(x) - f(0)|| le ||e_i||.||x|| le ||e_i||delta = ||e_i||fracepsilone_i = epsilon$.

I wonder if it's correct ?

real-analysis continuity vectors

share|cite|improve this question

edited 16 hours ago

Mathieu Rousseau

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. But this is not a definition of continuity, I'm afraid. It should go as something like: is continuous at $a$ when for each $epsilon>0$ there is $delta>0$ such that for all $xin A$ such that $lVert x-arVert<delta$ there holds $lVert f(x)-f(a)rVert<epsilon$.
  $endgroup$
  – user539887
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry I made a mistake in the definition. I corrected it.
  $endgroup$
  – Mathieu Rousseau
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is O.K. When you use the standard Euclidean norm, then $lVert e_irVert=1$. But perhaps by $lVertcdotrVert$ you mean any arbitrary norm, then $lVert e_irVert$ need not be $1$. Incidentally, Cauchy-Schwarz inequality, not inegality.
  $endgroup$
  – user539887
  19 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $f: mathbbR^m rightarrow mathbbR$. Show that $f(x) = x_i$ where $x_i$ is the $i^nd$ component of $x$ regarding the standard basis of $mathbbR^m$ is continuous in 0.

The definition says :

Let $A subset mathbbR^m$, $f: A rightarrow mathbbR^p$ and $a in A$. The function is continuous in a when for all $epsilon gt 0$, $exists delta gt 0$ such that for all $x in A$, $||x - a|| le delta$ and $|| f(x) - f(a) || le epsilon$.

I know $f(x) = (e_i|x)$.

By Cauchy-Schwarz inequality, we have : $|(e_i|x)| le ||e_i||.||x||$

We have $||x - 0|| = ||x|| le delta$.

So $||f(x) - f(0)|| le ||e_i||.||x|| - 0 = ||e_i||.||x||$

Now, for every $epsilon gt 0$ we can take $delta = fracepsilone_i$

So $||f(x) - f(0)|| le ||e_i||.||x|| le ||e_i||delta = ||e_i||fracepsilone_i = epsilon$.

I wonder if it's correct ?

real-analysis continuity vectors

share|cite|improve this question

edited 16 hours ago

Mathieu Rousseau

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 16 hours ago

Mathieu Rousseau

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 16 hours ago

Mathieu Rousseau

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Mathieu RousseauMathieu Rousseau

62