Let $H=L^2(a,b)$, and let $t_0$ be a fixed point in $(a,b)$. Let $f$ be the functional $H$ defined by $f(x)=x(t_0)$.
I wish to show that this functional is unbounded. I tried proof by contradiction but failed to get a contradiction. Is it actually bounded? Intuitively, it makes more sense to be unbounded. Either way, I am having trouble constructing a proof. Any help would be appreciated!
Edit: It is Example 3.7.2. in the photo below
As it was explained in the comments the functional is not well defined. But let's suppose the meaning was $C(a,b)$ with the norm of $L^2$. In that case the functional is well defined and really not bounded. Take a big enough $M>0$ such that $(t_0-\frac{1}{2M},t_0+\frac{1}{2M})\subseteq (a,b)$. Then define $x$ like this: $x$ is increasing linearly from $0$ to $M$ in $[t_0-\frac{1}{2M},t_0]$, then decreasing linearly from $M$ to $0$ in $[t_0,t_0+\frac{1}{2M}]$ and everywhere else it is zero. Then let $y=\sqrt{x}$, this is also a continuous function. Then obviously $|f(y)|=\sqrt{M}$ and $||y||_2=(\int_a^b |y|^2dt)^{\frac{1}{2}}=(\int_a^b xdt)^{\frac{1}{2}}=1$, this is clear from imagining the area below the graph in a geometric way. So $\sup_{y\ne 0}\frac{|f(y)|}{||y||_2}\geq \sqrt{M}$. This is true for any big enough $M$, so the supremum is not finite and the functional is not bounded.