Exercises in Sterling K. Berberian's Linear Algebra Recently I am reading Sterling K. Berberian's Linear Algebra but I got some questions on an exercise and I hope I could get some help here.
Here's the rephrased question:

Let $E$ be the inner product space in which vectors are continuous real-valued functions on a closed interval $[a, b]$ ($a$, $b$ are fixed constants) and the inner product is defined by integral: $<f,g>=\int_a^b f(t)g(t)dt$
Fix a point $c\in[a,b]$ and let $f$ be the linear form on $E$ defined by $f(x)=x(c)$ for all $x\in E$ (i.e. "pointwise evaluation" at $c$). Prove that there does not exist a function $y\in E$ such that $f(x)=<x,y> \forall x\in E$.
{Hint: Note that $f\neq 0$. Show that if $y\in E$, $y\neq 0$, then there exists $x\in E$ such that $x(c)=0$ but $\int_a^b x(t)y(t)dt>0$.}

The hint suggests me to pick $x$ as follow:
If $y(c)=0$, pick $x(t)=y(t)$.
If $y(c)>0$, pick $x(t)=\begin{cases}y(t)&,y(t)<0\\ 0&,y(t)\geq0\end{cases}$.
If $y(c)<0$, pick $x(t)=\begin{cases}y(t)&,y(t)>0\\ 0&,y(t)\leq0\end{cases}$.
But I think that this method of picking $x$ is unnatural, as it requires splitting into three different cases and using piecewise-defined functions. Is there some more naturally defined function which could also serve the same purpose? Thanks in advance.
 A: Suppose there exists a function $y\in E$ such that $f(x)=<x,y>$  $\forall x\in E$. Since $x+y \in E$, we have 
\begin{align*}
f(x+y) &= <x+y,y> \\
&= <x,y> + ||y||^2\,, 
\end{align*}
where $||y||^2 = \int_{a}^{b} y(t)^2 \,dt$.  
On the other hand since $f$ is a linear functional we have $f(x+y) = x(c)+y(c)$, and therefore the following should hold for any $x\in E$
\begin{equation*}
x(c)+y(c) = \int_a^b x(t)y(t)\,dt + ||y||^2 \quad (*)
\end{equation*}
But we can always choose $x(t)$ in such a way that the integral in $(*)$ goes to zero,  whereas LHS is zero and $||y||>0$, so we reach a contradiction! One such $x(t)$ is constructed below. 
$$ \Lambda (t) = 
\begin{cases} 1-\left| t-c \right|/\delta & \text{if} \hspace{3mm}|t-c|<\delta , \\ 
0 & \text{otherwise.}
\end{cases}$$
Note that $\Lambda(c)=1$ and $\Lambda(c+\delta)=\Lambda(c-\delta)=0$, so this is a triangular function with its tip at $(c,1)$ and base $(c-\delta,\,c+\delta)$. Choose 
\begin{equation*}
x(t)= -y(c) \Lambda(t)\,.
\end{equation*}
Therefore, $x(c)=-y(c)$ and $x(c+\delta)=x(c-\delta)=0$ as desired. Now let us see that the integral on the RHS of $(*)$ can be made arbitrarily small:
\begin{align*}
\left| \int_a^b x(t)y(t)\,dt \right| &= \left| \int_{c-\delta}^{c+\delta} x(t)y(t)\,dt \right| \\
&\leq  \int_{c-\delta}^{c+\delta} \left| x(t)\right| \left|y(t)\right| \,dt  \\
&\leq \int_{c-\delta}^{c+\delta} y(c) \bar{y} \,dt \quad \quad \left(\bar{y} = max_{t\in [a,b]} |y(t)| \right)\\
&= 2\delta y(c) \bar{y}\,.
\end{align*}
Finally, choose $\delta  = \frac{||y||^2}{4\delta y(c) \bar{y}}$. Then the RHS of $(*)$ is always greater than $||y||^2/2$, whereas LHS is zero. Contradiction!
A: I believe this is a better proof.
Assume the contrary, there exists such function $y(t)$. Pick $x(t)=(t-c)^2y(t)$ then $x(c)=0, x(t) \in E$ and we have $\int_{a}^b x(t)y(t) dt= \int_a^b \left[ (t-c)y(t) \right]^2dt= x(c)=0$, which leads to $(t-c)y(t)=0$ or $y(t)=0$. Hence, $\left \langle x,y \right \rangle =0=x(c)$ for all $x \in E$, which is a contradiction if we choose $x \in E$ so $x(c) \ne 0$.
