Show that $$\varphi:\ell^2\to\mathbb R, \varphi(x)=x(1)+x(2)$$ and $$\psi:\ell^1\to\mathbb R, \psi(x)=\sum\limits_{n=1}^\infty\frac{x(k)}{k}$$ are continuos and bounded
Linearity is straight forward we have $x,y\in\ell^2, \alpha\in\mathbb R$
$$\varphi(\alpha x+y)=\alpha (x(1)+x(2)) + y(1)+y(2) = \alpha\varphi(x)+\varphi(y)$$ and for $x,y\in\ell^1,\alpha\in\mathbb R$
$$\psi(\alpha x+y) = \sum\limits_{n=1}^\infty \frac{\alpha x(k)+y(k)}{k}=\alpha\sum\limits_{n=1}^\infty \frac{x(k)}{k}+\sum\limits_{n=1}^\infty \frac{y(k)}{k}=\alpha\psi(x)+\psi(y).$$
Now does it suffice to show that the operator norms $\|\varphi\|,\|\psi\|$ are bounded to show continuity?
The operator norm for $\varphi$ is $\|\varphi\|=\sup\limits_{\|x\|_{\ell^2}\leqslant 1}|\varphi(x)|=\sup\limits_{\|x\|_{\ell^2}=1}|\varphi(x)|$ furthermore $$\|x\|_{\ell^2}=\sqrt{\sum\limits_{n=1}^\infty|x(k)|^2} = 1 \iff \sum\limits_{n=1}^\infty|x(k)|^2 =1$$ from which follow that $\|\varphi\|=\sqrt2$ for $x = \left\lbrace\frac{\sqrt2}2,\frac{\sqrt2}2, 0, \dots\right\rbrace$.
The operator norm for $\psi$ is$\|\varphi\|=\sup\limits_{\|x\|_{\ell^1}\leqslant 1}|\psi(x)|=\sup\limits_{\|x\|_{\ell^1}=1}|\psi(x)|$ the $\ell^1$ norm is $1$ if $$\|x\|_{\ell^1}=\sum\limits_{n=1}^\infty|x(k)| = 1 \iff \sum\limits_{n=1}^\infty|x(k)| $$ because $\psi$ penalizes larger elements in the series more than earlier ones the operator norm is $\|\psi\|=1$ for example with $x = \left\lbrace1, 0, \dots\right\rbrace$.
Is this correct?