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I'm asked to prove that:

$$(\alpha_1, \alpha_2, \alpha_3,...) \mapsto \alpha_2$$

Is linear and continuous, where $( \alpha_1, \alpha_2, \alpha_3,... ) \in \ell^2$ and $\alpha_2 \in \mathbb{C}$. It is easy to prove that it is linear, but I'm stuck trying to show its continuity. By the definition:

$$\forall \varepsilon > 0, \exists \delta \ | \ \forall x \in D, ||x-x_0||<\delta \Rightarrow |f(x) - f(x_0)| < \varepsilon$$Thus:

$||x-x_0|| = \sqrt{(\alpha_1 - \alpha_1^0) (\alpha_1 - \alpha_1^0)^* + (\alpha_2 - \alpha_2^0) (\alpha_2 - \alpha_2^0)^* + (\alpha_3 - \alpha_3^0) (\alpha_3 - \alpha_3^0)^*} \ < \delta$, and $|f(x) - f(x_0)| = |\alpha_2 - \alpha_2^0| < \varepsilon$. I don't know how to find a relationship between $\delta$ and $\varepsilon$ to determine its continuity. Is the form non-continuous because of that?

FShrike
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ErikLAndre
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    Welcome. I made some changes to your formatting. You can use $$ to make a larger, displayed equation. You are also preferred to use, by general convention, the command \mapsto for the definition of a function mapping $x\mapsto f(x)$. The arrow \to is generally reserved for the arrow between domain and codomain, e.g. $\ell^2\to\Bbb C$. You also don't need \left,\right unless you've got something between the parantheses that enlarges them, like a fraction or an integral or some other large symbol – FShrike Oct 16 '22 at 11:40
  • Take $\delta =\epsilon$. – geetha290krm Oct 16 '22 at 11:45
  • Your linear form is $\alpha\mapsto\langle\alpha,\beta\rangle$ for $\beta=(0,1,0,0,\dots)$, and $\langle~,~\rangle$ is a continuous bilinear form. Note that by Riesz's representation theorem, you can apply this method to any candidate continuous linear form (it suffices to "guess" $\beta$). – Anne Bauval Oct 16 '22 at 11:49

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Since it is a linear function, to show continuity I highly recommend you use the following equivalent: there exists $M>0$, that $\|Tx\|\le M\|x\|$ for all $x\in\ell^2$. Or, even more nicely, for $\|x\|=1$ you just check that $\|Tx\|\le M$.

I define $T$ to be your linear map.

So let $\sum_{n=1}^\infty|\alpha_n|^2=1$. Of course, $|\alpha_n|\le1$ for every $n$. So quite simply, $\|T(\alpha_1,\cdots)\|=|\alpha_2|\le1$ regardless of the input. So $M=1$ suffices. And indeed you can check for yourself that the operator norm of $T$, $\|T\|$, is equal to $M=1$. $T$ is then continuous.

FShrike
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