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Consider the normed linear space $\Bbb R^2$ with this norm $\|(x,y)\|=|x|+|y|$.

Define $f$ to be a bounded linear functional on $X=\{(x,y):x=y\}$ by $f(x,y)=3x$.

If $g$ is a Hahn-Banach extension of $f$ on $\Bbb R^2$ given by $g(x,y)=\alpha x+\beta y$ then find $\alpha-\beta$.

My try:

Since $g$ is an extension of $f$ so $g|_X=f\implies g(x,x)=(\alpha+\beta)x=3x\implies \alpha+\beta =3$

Now $\|f\|=\sup \{f(x,x):\|(x,x)\|=1,(x,x)\in \Bbb R\}=3\implies \|g\|=3$

So $\sup \{g(x,y):\|(x,y)\|=1,(x,y)\in \Bbb R\}=3\implies \alpha x+\beta y\le 3$ where $|x|+|y|=1$.

Take $x=0,y=1\implies \alpha x+\beta y\le 3\implies \beta\le 3$.

How to find the value of $\alpha,\beta $ from above.

Please give some hints.

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    This seems to be quite frequent question, here is another copy: [What is the Hahn Banach extension of this map?](https://math.stackexchange.com/q/2130674) What is tne source of the problem? – 2017-12-25
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    Seems to be of GATE-2017 Examination. – 2017-12-29

1 Answers 1

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I think you have $\|f\| = \sup \{|3x| : 2|x| = \|(x,x)\| = 1\} = 3/2$, not $\|f\| =3$.

Then, since $\alpha + \beta =3$ and since $\max \{|\alpha |, |\beta|\} = 3/2$, you must have $\alpha = \beta = 3/2$ (why?). Hence, $\alpha - \beta =0$.

Here, I used that the dual norm to the $\ell^1$ norm is the $\ell^\infty $ norm.

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    Can't this be done without using duality? – 2017-02-28
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    Since $||g||=\frac{3}{2}\implies$ $\sup\{\alpha x+\beta y\}=\frac{3}{2};|x|+|y|=1$ The extrema points of $|x|+|y|=1=(\pm 1,0),(0,\pm 1)$ – 2017-02-28
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    Hence at $(1,0)$ we have $\alpha x+\beta y\implies \alpha=\frac{3}{2}$;Similarly at $(0,1)$ we have $\alpha x+\beta y\implies \beta=\frac{3}{2}\implies \alpha=\beta$ – 2017-02-28
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    Is it correct?Thanks for pointing $||f||=\frac{3}{2}$ – 2017-02-28
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    @Ben: In the present case, the formula for the dual is very easy to see (this is essentially what you are doing yourself): It is easy to see $\|(x,y)\|_{(\ell^1)'} \leq \|(x,y)\|_{\ell^\infty } $. Conversely, $\|(x,y)\|_{(\ell^1)'} \geq |x|$, as can be seen by "testing" with $(1,0) $. Testing with $(0,1) $ completes the proof. – 2017-02-28