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**Question 1:**

Consider $V = \mathbb R$ as a vector space over itself, and let $K := [0, 0.5] \cap [2, 2.5]$, and suppose the Hyperplane seperation theorem were to hold. Then since $1.5$ is not in $K$, there must be some nonzero linear functional $f$ such that $f(1) = c$ for some $c \in \mathbb R$, and $f(x) \leq c$ for all $x \in K$.

We first note that by scaling, we have $f(x) = xf(1)$ for all $x \in \mathbb R$.

This means in particular that $c \neq 0$, for otherwise we would have $ f(x) = xf(1) = 0$ for all $ x \in \mathbb R$, i.e. $f$ must be the zero functional.

Assume now that $c > 0$. Then $f(2) = 2c  > c$, which contradicts the Hyperplane seperation theorem. On the other hand if $c < 0$ we get similarly the contradiction $f(0.5) = 0.5c > c$.

Thus the Hyperplane seperation theorem does not hold for this $K$.

**Question 2:**

We denote $p' = q$ for convenience.

$\ell$ is a bounded functional on $L^p$. Indeed, let $g = \mathbf 1_{[1/3, 1/2]} - \mathbf 1_{[3/4, 1]}$, where $\mathbf 1_U$ denotes the indicator function of the set $U$. Then $g \in L^{q}$ since the $L^q$ norm of $g$ is $(\int_{0}^1 |g|^q \ d\mu)^{\frac{1}{q}} = (\int_{1/3}^{1/2} 1 \ d\mu + \int_{3/4}^{1} 1 \ d\mu)^{\frac {1}{q}} = (\frac{5}{12})^{\frac{1}{q}}$. This answers part (c).

We further note that we have that for all $f \in L^p$, $\ell(f) = \int_{0}^1 fg \ d\mu$ - indeed, we have $\int_{0}^1 fg \ d\mu = \int_{0}^1 f \mathbf 1_{[1/3, 1/2]} - f \mathbf 1_{[3/4, 1]} = (\int_{1/3}^{1/2} f \ d\mu + \int_{3/4}^{1} f \ d\mu) = \ell(f)$. This answers part (b) affirmatively.

By Holder's inequality, we have that $\int_0^1 fg \ d\mu \leq \|f\|_{p} \|g \|_q$ so that $\frac{\|l(f)\|}{\|f\|_p} \leq \|g\|_q = (\frac{5}{12})^{\frac{1}{q}}$. Taking the sup over $f$, we conclude that $\|l\| \leq (\frac{5}{12})^{\frac{1}{q}}$ (Equation 1), in particular $\ell$ is bounded. Thus we answer the first portion of part (a) affirmatively.

I claim that in fact $\|l\| = (\frac{5}{12})^{\frac{1}{q}}$. To see this, take $f = g = \mathbf 1_{[1/3, 1/2]} - \mathbf 1_{[3/4, 1]}$.

Then $\|f\|_p =$

$(\int_{1/3}^{1/2} 1 \ d\mu + \int_{3/4}^1 1 \ d\mu)^{\frac {1}{p}}$

$= (\frac{5}{12})^{\frac{1}{p}}$.

While $\ell(f) =  \int_0^1 g^2 \ d\mu = (\int_{1/3}^{1/2} 1 \ d\mu + \int_{3/4}^{1} 1 \ d\mu) = \frac{5}{12}$

Whence $\frac{\|\ell (f)\|}{\|f\|_p} = (\frac{5}{12})^{1 - \frac{1}{p}} = (\frac{5}{12})^{\frac{1}{q}}$. Combining this with Equation 1, we obtain the desired result. This answers the second portion of part (a).

**Question 3:**

Let $x = (x_1, x_2, \dots) \in c_0$ and $y = (y_1, y_2, \dots) \in \ell_1$. Define $\psi_y: c_0 \to \mathbb R$ by $\psi_y (x) = \sum_{k = 1}^{\infty} x_k y_k$. I claim that $\psi_y$ is a bounded linear functional.

Indeed, note that since $x \in c_0$, we have $x \in \ell^{\infty}$ also, so that by Holder's inequality for $\ell_1$ and $\ell_{\infty}$ we have

$\sum_{k = 1}^{\infty} |x_k y_k| \leq \lvert \lvert x \rvert \rvert_{\infty} \lvert \lvert y \rvert \rvert_1$. (Equation 1)

Thus $\sum_{k = 1}^{\infty} x_k y_k$ converges absolutely, and so for $v, w \in c_0$, we have

$\psi_y (v + w) = \sum_{k = 1}^{\infty} (v_k + w_k)(y_k) =  \sum_{k = 1}^{\infty} v_k y_k + \sum_{k = 1}^{\infty} w_k y_k = \psi_y (v) + \psi_y (w),$

and for any $\lambda \in \mathbb R$, we have $\psi_y (\lambda x) = \sum_{k = 1}^{\infty} \lambda x_k = \lambda \sum_{k = 1}^{\infty} x_k = \lambda \psi_y (x)$.

So $\psi$ is linear.

Further, from Equation 1 we have

$\lvert \lvert \psi_y (x) \rvert \rvert = \lvert \sum_{k = 1}^{\infty} x_k y_k \rvert \leq \sum_{k = 1}^{\infty }\lvert x_k y_k \rvert \leq \lvert \lvert x \rvert \rvert_{\infty} \lvert \lvert y \rvert \rvert_{1}$

so that

$\lvert \lvert \psi_y \rvert \rvert \leq \lvert \lvert y \rvert \rvert_{1}$. (Equation 2)

Hence $\psi$ is also bounded as claimed.

Now I claim that every bounded linear functional is of the form $\psi_y$ for some $y \in \ell^1$. Let $\phi$ thus be an arbitrary bounded linear functional.

We denote by $e^i$ the sequence with a $1$ in the $i$'th slot, and $0$ otherwise.

For each positive integer $N$, define $q_N := \sum_{k = 1}^{N} \text{sgn}(\phi(e^k)) e^k$, where by convention, $\text{sgn}(0) = 0$. It is clear that $\lvert \lvert q_N \rvert \rvert_{\infty} = 1$ and $q_N \in c_0$.

We compute

$\phi(q_N) = \phi (\sum_{k = 1}^{N} \text{sgn}(\phi(e^k)) e^k) = \sum_{k = 1}^{N} \text{sgn}(\phi(e^k)) \phi(e^k) = \sum_{k = 1}^{N} |\phi(e_k)| \leq \lvert \lvert \phi \rvert \rvert$.

where the third equality is by linearity of $\phi$, and the fifth by the fact that $\lvert \lvert q_N \rvert \rvert_{\infty} = 1$.

Taking the limit as $N \to \infty$ of the above, we have

$\sum_{k = 1}^{\infty} |\phi(e^k)| \leq \lvert \lvert \phi \rvert \rvert$ (Equation 3), hence the sequence $y$ defined by $y_k = \phi(e^k)$ is in $\ell_1$.

I claim that this is the $y$ we seek. Indeed, for sequences $x \in c^f_0$, the set of sequences with only finitely many elements nonzero, we have by linearity that

$\phi(x) = \phi (\sum_{i = 1}^{j} x_i e^i ) = \sum_{i = 1}^{j}x_i \phi (e^i) = \sum_{i = 1}^{j}x_i y_i = \sum_{i = 1}^{\infty}x_i y_i$.

so that the formula holds at least for all such sequences.

Now $c^f_0 \subset c_0$ and is even dense in $c_0$ - indeed, given $x \in c_0$ define the sequence $w^n := (x_1, \dots , x_n, 0, \dots)$. Then $w^n \in c^f_0$ and $w^n \to x$ in $\ell_{\infty}$ norm.

We conclude thus by continuity of $\phi$ that the formula $\phi(x) = \sum_{i = 1}^{\infty}x_i y_i$ holds for all $x \in c_0$. This concludes our earlier claim.

Thus we have established a bijection between $\ell_1$ and the bounded linear functionals on $c_0$ - that is, $y \to \psi_y$.

It remains only to show that $y \to \psi_y$ is an isometry. But $\lvert \lvert \psi_y \rvert \rvert \leq \lvert \lvert y \rvert \rvert_1$ is Equation 2, and $\lvert \lvert y \rvert \rvert_1 \leq \lvert \lvert \psi_y \rvert \rvert$ is Equation 3, so $\lvert \lvert y \rvert \rvert_1 = \lvert \lvert \psi_y \rvert \rvert$ and the map is an isometry as claimed.

That concludes the proof.

**Question 4:**

a) We compute $|\ell(f)| = |f(\frac{2}{3})| \leq \sup_{x \in [0, 1]} |f(x)| = \lvert \lvert f \rvert \rvert_{C[0, 1]}$, that is, $\frac{\lvert \lvert \ell (f) \rvert \rvert}{\lvert \lvert f \rvert \rvert_{C[0, 1]}} \leq 1$ Taking the supremum over $f$, we have $\lvert \lvert \ell \rvert \rvert \leq 1$.

To show that we actually have equality in the above, i.e. $\lvert \lvert \ell \rvert \rvert = 1$, take $f$ to be the constant function equal to $1$ everywhere. Then $f(\frac{2}{3}) = \lvert \lvert f \rvert \rvert_{C[0, 1]} = 1$, so the supremum is achieved $\lvert \lvert \ell \rvert \rvert = 1$ as desired.

b) We prove by contradiction. Suppose there existed such a $g \in L^1 [0, 1]$. Consider the sequence of functions $f_n \in C[0, 1] \subset L^{\infty}$ defined for each $n \geq 1$ as follows:

At the points $0, \frac{2}{3} - \frac{1}{2^{n+1}}, \frac{2}{3}, \frac{2}{3} + \frac{1}{2^{n+1}}, 1$, we define $f$ by

$f_n (0) = f_n ( \frac{2}{3} - \frac{1}{2^{n+1}}) = f(\frac{2}{3} + \frac{1}{2^{n+1}}) = f_n(1) = 0$; $f_n(\frac{2}{3}) = 1$.

For all other points, we linearly interpolate between the values of $f$ at these five points.

Clearly $f_n \in C[0, 1]$ for each $n$. Further, we have  $|\ell(f_n)| = f_n (\frac{2}{3}) = 1$ for each $n$. However, we compute

Thu, 07 Oct 2021 12:42 GMT