Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$












4












$begingroup$


Can you check if my proof is correct?




Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$




Let $fin X^{*}$, then $f$ is a bounded linear functional. Assume that $x,in X$ such that begin{align}f(x)=f(y)&iff f(x)-f(y)=0, \& iff f(x-y)=0, ;text{since};f ;text{is a linear functional};\& iff x-yin ker f ={0}\& iff x=yend{align}






linear-algebra functional-analysis analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:25










  • $begingroup$
    @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:27












  • $begingroup$
    @Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:29












  • $begingroup$
    LOL, What if $X=Bbb R^n$? What's the kernel then?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:29






  • 1




    $begingroup$
    Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:31
















4












$begingroup$


Can you check if my proof is correct?




Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$




Let $fin X^{*}$, then $f$ is a bounded linear functional. Assume that $x,in X$ such that begin{align}f(x)=f(y)&iff f(x)-f(y)=0, \& iff f(x-y)=0, ;text{since};f ;text{is a linear functional};\& iff x-yin ker f ={0}\& iff x=yend{align}






linear-algebra functional-analysis analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:25










  • $begingroup$
    @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:27












  • $begingroup$
    @Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:29












  • $begingroup$
    LOL, What if $X=Bbb R^n$? What's the kernel then?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:29






  • 1




    $begingroup$
    Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:31














4












4








4


2



$begingroup$


Can you check if my proof is correct?




Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$




Let $fin X^{*}$, then $f$ is a bounded linear functional. Assume that $x,in X$ such that begin{align}f(x)=f(y)&iff f(x)-f(y)=0, \& iff f(x-y)=0, ;text{since};f ;text{is a linear functional};\& iff x-yin ker f ={0}\& iff x=yend{align}






linear-algebra functional-analysis analysis







share|cite|improve this question











$endgroup$




Can you check if my proof is correct?




Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$




Let $fin X^{*}$, then $f$ is a bounded linear functional. Assume that $x,in X$ such that begin{align}f(x)=f(y)&iff f(x)-f(y)=0, \& iff f(x-y)=0, ;text{since};f ;text{is a linear functional};\& iff x-yin ker f ={0}\& iff x=yend{align}






linear-algebra functional-analysis analysis




linear-algebra functional-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 19:26







Omojola Micheal

















asked Jan 10 at 19:23









Omojola MichealOmojola Micheal

1,831324




1,831324








  • 1




    $begingroup$
    Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:25










  • $begingroup$
    @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:27












  • $begingroup$
    @Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:29












  • $begingroup$
    LOL, What if $X=Bbb R^n$? What's the kernel then?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:29






  • 1




    $begingroup$
    Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:31














  • 1




    $begingroup$
    Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:25










  • $begingroup$
    @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:27












  • $begingroup$
    @Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:29












  • $begingroup$
    LOL, What if $X=Bbb R^n$? What's the kernel then?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:29






  • 1




    $begingroup$
    Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:31








1




1




$begingroup$
Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
$endgroup$
– Ted Shifrin
Jan 10 at 19:25




$begingroup$
Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
$endgroup$
– Ted Shifrin
Jan 10 at 19:25












$begingroup$
@Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
$endgroup$
– Omojola Micheal
Jan 10 at 19:27






$begingroup$
@Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
$endgroup$
– Omojola Micheal
Jan 10 at 19:27














$begingroup$
@Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
$endgroup$
– Omojola Micheal
Jan 10 at 19:29






$begingroup$
@Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
$endgroup$
– Omojola Micheal
Jan 10 at 19:29














$begingroup$
LOL, What if $X=Bbb R^n$? What's the kernel then?
$endgroup$
– Ted Shifrin
Jan 10 at 19:29




$begingroup$
LOL, What if $X=Bbb R^n$? What's the kernel then?
$endgroup$
– Ted Shifrin
Jan 10 at 19:29




1




1




$begingroup$
Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
$endgroup$
– Ted Shifrin
Jan 10 at 19:31




$begingroup$
Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
$endgroup$
– Ted Shifrin
Jan 10 at 19:31










2 Answers
2






active

oldest

votes


















4












$begingroup$

The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10



















2












$begingroup$

Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069076%2fprove-that-if-fx-fy-for-all-f-in-x-then-x-y%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10
















4












$begingroup$

The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10














4












4








4





$begingroup$

The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).






share|cite|improve this answer









$endgroup$



The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 19:36









triple_sectriple_sec

15.8k21851




15.8k21851












  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10


















  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10
















$begingroup$
Thanks for your quick response. I'll read through (+1)
$endgroup$
– Omojola Micheal
Jan 10 at 19:38




$begingroup$
Thanks for your quick response. I'll read through (+1)
$endgroup$
– Omojola Micheal
Jan 10 at 19:38












$begingroup$
I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
$endgroup$
– Omojola Micheal
Jan 10 at 19:48






$begingroup$
I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
$endgroup$
– Omojola Micheal
Jan 10 at 19:48














$begingroup$
The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
$endgroup$
– triple_sec
Jan 10 at 20:05






$begingroup$
The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
$endgroup$
– triple_sec
Jan 10 at 20:05














$begingroup$
Yes, that's true!
$endgroup$
– Omojola Micheal
Jan 10 at 20:10




$begingroup$
Yes, that's true!
$endgroup$
– Omojola Micheal
Jan 10 at 20:10











2












$begingroup$

Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37
















2












$begingroup$

Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37














2












2








2





$begingroup$

Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}






share|cite|improve this answer









$endgroup$



Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 19:34









José Carlos SantosJosé Carlos Santos

156k22125227




156k22125227












  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37


















  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37
















$begingroup$
Thanks for your quick response. (+1)
$endgroup$
– Omojola Micheal
Jan 10 at 19:37




$begingroup$
Thanks for your quick response. (+1)
$endgroup$
– Omojola Micheal
Jan 10 at 19:37


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069076%2fprove-that-if-fx-fy-for-all-f-in-x-then-x-y%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more