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The norm on $\mathbb{R}^2$ is defined as follows: $$ \| (x,y) \|=|x|+|y|. $$ Now let $$X=:\{ (x,x):\ x\in \mathbb{R}\}$$ be a subspace of $\mathbb{R}^2. $ Let $ f:X\to \mathbb{R}$ defined as $f(x,y)=3x,\ \forall (x,y)\in X.$ Let $g$ be an extension of $f$ defined as $g(x,y)=\alpha x+\beta y,\ \forall (x,y)\in \mathbb{R}^2$. Then what is the value of $\alpha $ and $\beta.$

Since, $g$ is an extension of $f$, so for every $(x,y)\in X$, \begin{align*} g(x,y)& =f(x,y)=3x\\ \implies \alpha x+\beta y&=3x\\ \implies \alpha x+\beta x& =3x \tag{$x=y$ on $X$}\\ \implies \alpha +\beta &=3 \end{align*} Now the Hahn Banach extension preserves the norm of the functional, i.e. $\|g\|=\|f\|$. The norm of $f$ is, \begin{align*} \|f\| & = \sup\left\lbrace \frac{|f(x,x)|}{\|(x,x)\|}:\ (x,x)\neq 0 \right\rbrace \\ & = \sup\left\lbrace \frac{|3x|}{|x|+|x|}:\ x\neq 0 \right\rbrace\\ & = \sup\left\lbrace \frac{3}{2}:\ x\neq 0 \right\rbrace\\ & = \frac{3}{2} \end{align*} Now $\|g\|=\|f\|=\frac{3}{2}$ \begin{align*} \|g\| & = \sup\left\lbrace \frac{|g(x,y)|}{\|(x,y)\|}:\ \|(x,y)\|\neq 0 \right\rbrace\\ & = \sup\left\lbrace \frac{|\alpha x+\beta y|}{\|(x,y)\|}:\ \|(x,y)\|\neq 0 \right\rbrace\\ & = \sup\left\lbrace \frac{|(3-\beta) x+\beta y|}{\|(x,y)\|}:\ \|(x,y)\|\neq 0 \right\rbrace\\ & = \sup\left\lbrace \frac{|3x+\beta(y-x)|}{|x|+|y|}:\ \|(x,y)\|\neq 0 \right\rbrace\\ \end{align*} After that I stuck. Can anyone help me please?

Thanks.

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Hint: Note that $$|g(x,y)|\leq\frac{1}{2}\max\{|\alpha|,|\beta|\}\|(x,y)\|.$$ Then try to show that this is a tight bound, i.e. find a $(x,y)$ which gives you equality. By showing this, you will get $\|g\|=\frac{1}{2}\max\{|\alpha|,|\beta|\}$. Finally use what you have shown already to determine $\alpha,\beta$.

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  • $\begingroup$ $|g(x,y)| = |\alpha x + \beta y| \leq \max\{|\alpha|,|\beta|\}(|x|+|y|)$. Where does the $\frac{1}{2}$ come from? (eg, $||(1,0)|| = 1$, and $||g(1,0)|| = |\alpha|$) $\endgroup$ – Christian Fieldhouse Apr 1 '18 at 11:55

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