"Let $\mathcal H$ be a complex Hilbert space and let $y\in\mathcal H.$ Show that the linear transformation $f:\mathcal H\to\mathbb C$ defined by, $f(x)=\langle x,y\rangle$ is continuous."
Here is what I have done:
$$|f(x)|=|\langle x,y\rangle|$$
$$\le\|x\|\|y\|,\,\,\,\,\forall\, x\in\mathcal H$$
By the Cauchy-Schwarz inequality.
Since $y\in\mathcal H$ is fixed, I was going to say that $f$ is continuous (bounded) with $k=\|y\|$, where $k$ is the constant in the definition of a bounded linear operator,
$$\|Tx\|\le k\|x\|,k\gt0$$
For some linear operator $T$, in this case $f$.
In the solution, however, it starts off with $|f(x)|^2$ as opposed to just $|f(x)|$ in my attempt. Why do you start off by considering $|f(x)|^2$? Clearly both ways don't yield the same answer, as the latter yields $k=\|y\|^2$.
I can see that taking the square of $|f(x)|$ allows us to relate to the norm induced by the inner product, namely, that $\|x\|=\sqrt{\langle x,x\rangle}$, but still am not entirely sure on why one would need to do that.
It´s right: The Cauchy -Schwarz inequality $|f(x)|\leq ||x|| ||y||$ shows that $||f||\leq ||y||$ where $||f||$ denotes the operator norm and since for $y\neq0$ You have $f(\frac{y}{||y||})=||y||$ it follows $||f||=||y||$. If " the solution" yields $||f||=||y||^2$ there must be a mistake.