## Recommended Functional Analysis Book (Graduate Level)

Functional Analysis is a subject that combines Linear Algebra with Analysis. I researched online, and it seems one of the best Functional Analysis Book for Graduate level is Functional Analysis by Peter Lax. This book is ideal for a second course in Functional Analysis. For a first course in Functional Analysis, I would recommend Kreyszig, which is listed on my Recommended Undergraduate Math Books page.

This is Theorem 5 in the book: Let $X$ be a linear space over the reals.

The following 9 properties hold:

1. The empty set is convex.
2. A subset consisting of a single point is convex.
3. Every linear subspace of $X$ is convex.
4. The sum of two convex subsets is convex.
5. If $K$ is convex, so is $-K$.
6. The intersection of an arbitrary collection of convex sets is convex.
7. Let $\{K_j\}$ be a collection of convex subsets that is totally ordered by inclusion. Then their union $\cup K_j$ is convex.
8. The image of a convex set under a linear map is convex.
9. The inverse image of a convex set under a linear map is convex.

### Brief sketch of proofs:

We give a brief sketch of the idea behind the proofs.

We are using the definition of convex as follows: $X$ is a linear space over the reals; a subset $K$ of $X$ is called convex if, whenever $x$ and $y$ belong to $K$, all points of the form $ax+(1-a)y$, $0\leq a\leq 1$ also belong to $K$.

Property 1 is vacuously true.

Property 2 is true because of $ax+(1-a)x\equiv x$.

Property 3 is true because $ax+(1-a)y$ is a linear combination and is thus in the linear subspace.

Property 4) Let $C_1$ and $C_2$ be the two convex subsets. Let $x_1+y_1$ and $x_2+y_2$ be points in $C_1+C_2=\{x+y:x\in C_1, y\in C_2\}$

\begin{aligned} a(x_1+y_1)+(1-a)(x_2+y_2)&=ax_1+ay_1+x_2+y_2-ax_2-ay_2\\ &=[ax_1+(1-a)x_2]+[ay_1+(1-a)y_2]\\ &\in C_1+C_2 \end{aligned}

Property 5) We just need to know that $-K=\{-x:x\in K\}$ and this algebraic observation: $a(-x)+(1-a)(-y)=-(ax+(1-a)y)$.

Property 6) Let $x,y\in\bigcap_{i\in I}C_i$. $ax+(1-a)y\in C_i$ for all $i\in I$, thus $ax+(1-a)y\in\bigcap_{i\in I}C_i$.

Property 7) Let $x,y\in\bigcup K_j$, where $K_i\subseteq K_{i+1}$. Let $x\in K_n$, $y\in K_m$, then either $K_n\subseteq K_m$ or $K_m\subseteq K_n$. If $K_n\subseteq K_m$, $ax+(1-a)y\subseteq K_m\subseteq \bigcup K_j$. Similarly for the other case $K_m\subseteq K_n$.

Property 8) Observe that $af(x)+(1-a)f(y)=f(ax+(1-a)y)\in f(K)$.

Property 9) The only tricky thing about this part is that we cannot assume that the inverse $f^{-1}$ exists. We can only talk about the pre-image.

Let $w,z\in f^{-1}(K)$. $f(w)\in K$ and $f(z)\in K$.

We have $f(aw+(1-a)z)=af(w)+(1-a)f(z)\in K$.

Thus $aw+(1-a)z\in f^{-1}(K)$.

The End!