Here is the first question:
Let $X$ and $Y \neq \{0\}$ be normed spaces, where $dim X = \infty.$ Show that there is at least one unbounded linear operator $T: X \rightarrow Y.$ (Use a Hamel basis.)
Here is the second question:
Let $X$ be a normed linear space that is not finite dimensional. Prove that there exists an unbounded linear functional on $X.$
Are those 2 questions asking about the same thing? If not how they are different?
I am asking this because I know the answer to the first question and I want to use it to answer the second question.
EDIT:
Here is the solution to the first question:
Solution
Let $H$ be an Hamel basis for $X$ (see theorem $4.1-7$ ). We will show that any function $T: H \rightarrow Y$ can be extended to a linear operator $\tilde{T}: X \rightarrow Y$ Let $x$ be in $X .$ since $H$ is an Hamel basis, $\exists x_{1}, \ldots, x_{n}$ in $H$ and scalars $\alpha_{1}, \ldots, \alpha_{n}$ such that $x=\alpha_{1} x_{1}+\ldots+\alpha_{n} x_{n} .$ So define $$ \tilde{T}(x)=\alpha_{1} T\left(x_{1}\right)+\ldots+\alpha_{n} T\left(x_{n}\right) $$ Definition is well-defined since any element $x$ in $X$ has unique representation wrt $H . \tilde{T}$ is linear because if $x, y \in X$ and $\alpha, \beta$ are scalars (say $x=\alpha_{1} x_{1}+\ldots+\alpha_{n} x_{n}$ and $\left.y=\beta_{1} y_{1}+\ldots+\beta_{k} y_{k} \text { wrt } H\right)$ then $$ \begin{aligned} \tilde{T}(\alpha x+\beta y) &=\tilde{T}\left(\alpha\left(\alpha_{1} x_{1}+\ldots+\alpha_{n} x_{n}\right)+\beta\left(\beta_{1} y_{1}+\ldots+\beta_{k} y_{k}\right)\right) \\ &=\tilde{T}\left(\alpha \alpha_{1} x_{1}+\ldots+\alpha \alpha_{n} x_{n}+\beta \beta_{1} y_{1}+\ldots+\beta \beta_{k} y_{k}\right) \\ &=\alpha \alpha_{1} T\left(x_{1}\right)+\ldots+\alpha \alpha_{n} T\left(x_{n}\right)+\beta \beta_{1} T\left(y_{1}\right)+\ldots+\beta \beta_{k} T\left(y_{k}\right) \\ &=\alpha\left[\alpha_{1} T\left(x_{1}\right)+\ldots+\alpha_{n} T\left(x_{n}\right)\right]+\beta\left[\beta_{1} T\left(y_{1}\right)+\ldots+\beta_{k} T\left(y_{k}\right)\right] \\ &=\alpha \bar{T}(x)+\beta \bar{T}(y) \end{aligned} $$ Lets turn the question. since $H$ is an infinite set let $v_{1}, v_{2}, \ldots$ be in $H .$ Fix non-zero $y \in Y .$ Define the function $T: H \rightarrow Y$ by $T\left(v_{n}\right)=n\left\|v_{n}\right\| y$ and $\mathbf{0}$ otherwise. Then $T$ can be extended to a linear operator $\tilde{T}: X \rightarrow Y .$ But $\tilde{T}$ can not be bounded, given $c \in \mathbb{R}_{+}$ choose $n>c /\|y\|$, so $\left\|\tilde{T} v_{n}\right\|=n\left\|v_{n}\right\|\|y\|>c\left\|v_{n}\right\|$
The second question is different from the first one, since a linear functional on $X$ is a linear operator from $X$ into $\mathbb R$ (or $\mathbb C$). So, in principle, there might be such an example for the first question, and not to exist an example for the second one (that's not the case though).