Let's consider the following space:
$$ X = \{f \in C^\infty [0,1]: f(0)=f(1)=0 \}$$
with an inner product given by
$$ \langle f,g \rangle = \int_0^1 f'(t) \overline{g'(t)}dt. $$
Moreover, let's fix $h \in C^\infty [0,1]$ and define operator $M_h:X \rightarrow X$ given by $$M_h(f) = h(t)f(t)$$ for $f \in X$ and $t \in [0,1]$.
First question is to show that space $X$ with this inner product is not a Hilbert space (motivation being that the completion of this space is the Sobolev space $H^1_0[0,1]$). My idea was the following: I wanted to find a sequence of smooth functions similar to the bump function https://en.wikipedia.org/wiki/Bump_function , for example given by $$ f(t) = \begin{cases} \operatorname{exp}\Big(\frac{-(t-\frac{1}{2})^2}{\frac{1}{2}^2 - (t-\frac{1}{2})^2}\Big), & t \in (0,1) \\ 0, & t \in \{0,1\} \end{cases}$$
such that the mass becomes more and more concentrated in the middle (but the area of the integral of the differences keeps being small, so it would be a Cauchy sequence), and this sequence would converge to a non-continuous (but integrable) function. Unfortunately I'm not sure if this is a reasonable approach or how to explicitly describe this, since I feel like functions from the space $X$ have very strong properties so it's hard to find a sequence "breaking" it. I would appreciate any feedback about it, if possible.
The second question is to show that operator $M_h$ is bounded. I managed to show this (I will write $h$ and $f$ instead of $h(t)$ and $f(t)$ for better clarity): denote $M = \max\{M_1, M_2\}$ where $$ M_1 = \max_{t \in [0,1]} |h(t)|$$ $$ M_2 = \max_{t \in [0,1]} |h'(t)|.$$
Then \begin{align} \Vert M_h \Vert^2 &= \Vert hf \Vert = \int_0^1 (hf)'\overline{(hf)'}dt =\int_0^1 |(hf)'|^2 \\&= \int_0^1 |h'f + hf'|^2 \leq \int_0^1 (|h|^2+|h'|^2)(|f|^2 + |f'|^ 2 )dt \\ &\leq 2M^2 \Big( \int_0^1 |f|^2 dt + \int_0^1 |f'|^2 dt \Big) \\ &=2M^2 \Big( \int_0^1 |f|^2 dt + \Vert f \Vert ^2 \Big) \end{align} where the first inequality comes from Cauchy-Schwartz lemma. Now, I would like to do something about the remaining integral, namely $\int_0^1 |f|^2 dt$ but here I'm stuck. Could someone share a hint if this approach makes sense, assuming my previous calculations were correct?
Thanks!
You wrote $ \int_0^1 |f|^2 dt + \| f\|$ for $ \int_0^1 |f|^2 dt + \| f\|^{2}$
$f(x)=\int_0^{x} f'(t)dt$ so $|f(x)| \leq \int_0^{1}|f'(t)| dt\leq \sqrt {\int_0^{1}|f'(t)|^{2} dt}$ This implies that $\int_0^{1}|f(t)|^{2} dt \leq \int_0^{1}|f'(t)|^{2} dt=\|f\|^{2}$.
Here is a different approach to the first part. Define $T(f)=f(\frac 1 2)=\int_0^{\frac 1 2} f'(t)dt$. This is clearly a continuous linear functional. If the space is complete then we can apply Riesz Theorem to conclude that there exists $g \in X$ with $Tf=\int_0^{1}f'(t)g'(t)dt$ for all $f \in X$. Thus $\int_0^{1} f'(t)[g'(t)-\chi_{(0,\frac 1 2)}] dt=0$ for all $f \in X$. This implies that $g'(t)=\chi_{(0,\frac 1 2)}$ a.e. but this contradicts the fact that $g'$ is continuous.